% The following TeXfile comprises part of

%     MEASURE THEORY, by D.H.Fremlin
%          (University of Essex, Colchester, England;
%                   e-mail  fremdh@essex.ac.uk)

% Please contact me with any difficulties not covered by the remarks
% below.

% The file is set up for processing by plain TeX in the presence of
% files  amstex.tex ,  amssym.tex  and  amsppt.sty  suitably supported by the
% appropriate fonts.   If these are unavailable, a readable version
% can be obtained by removing the comment symbol  %  from the line

%        %   \amstexsupportedfalse

% below.   Some systems do not allow access to files  amstex.tex  and
% amsppt.sty , but instead employ a command "amstex <filename.tex>";
% for such systems it is sometimes right to delete or disable the
% commands

%        \input amstex
%        \documentstyle{amsppt}

% (I have little experience with this and should welcome comments.)

% Various formats of output are obtainable.   The principal variations
% are

%    1. "Standard" output uses the command
%               \smallprint
% and gives output adapted to A4 paper (21 cm x 29.7 cm).

%    2.  Using the command
%               \USsmallprint
% gives output adapted to 8.5" x 11" .

%    3.  Using the command
%               \noproofs
% produces an abridged version, in which proofs are omitted.

%    4.   Using the command
%               \resultsonly
% produces a super-abridged version, omitting comments and exercises
% as well as proofs.

% These commands are grouped together at the end of the introductory
% definitions;  they may be activated by deleting the initial comment
% symbols  % .

% Any diagrams included will be called for by inputting postscript files.
% (You can look for these by hunting for   \picture  in the code below.)
% To get these you will need the relevant  .ps  files with the expected
% names  ( \picture{mt133g}{...}  demands the file  mt133g.ps , for instance).


%(If you can read TeX fluently, you may wish to fast-forward to a line
%beginning   %file starts here   or   \frfilename
%if one of these is present.)

%TeXfiles with this heading are designed for processing with AmSTeX,
%accessed through TeXfiles  amstex.tex (with amssym.tex  and
% amsppt.sty  and possibly  amsppt.sti)  during compilation.   
%If these are unavailable, a readable
%version can (I hope) be produced by deleting the  %  before
% \amstexsupportedfalse  below.

%Some special difficulties arise with the importation of  .ps  files
%in diagrams.   Look for  \picture  below for notes on these.

%The following \count registers are used for special purposes:
%   \count1=\stylenumber  to specify style
%   \count2=\footnotenumber  to count footnotes

%AmSTeX declaration
\newif\ifamstexsupported
     \amstexsupportedtrue
%     \amstexsupportedfalse               %use when AMSTeX unavailable
\newif\ifDHFamstexloaded        %for detection of a non-standard AMSTeX
     \DHFamstexloadedfalse    
\newif\ifDHFamspptloaded        
     \DHFamspptloadedfalse    
\newif\iffourteensavailable
     \fourteensavailabletrue
    \fourteensavailablefalse
\newif\ifathome
     \athometrue     %to cope with a quirk     of my son's Ghostscript
     \athomefalse
\newif\iffirstpage
     \firstpagetrue

%Style assignments
\countdef\stylenumber=1
\countdef\footnotenumber=2
\def\fullsize{\magnification=\magstep1
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=1}
\def\smallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=2}
\def\printersize{\magnification=\magstep1
     \baselineskip24pt
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{8.5truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=3}
\def\refsize{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{390pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=4}
\def\rrversion{\magnification=\magstep1
     \overfullrule0pt
     \def\pagewidth{330pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=5}
\def\bigsize{\magnification=\magstep2
     \overfullrule0pt
     \def\pagewidth{325pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=6}
\def\bbigsize{\magnification=\magstep3
     \overfullrule0pt
     \def\pagewidth{271pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=7}
\def\USsmallprint{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{482pt}
     \def\pageheight{8.7truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=8}
\def\halfmagsize{\magnification=\magstephalf
     \def\pagewidth{427pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \stylenumber=9}
           % register  stylenumber  is used to control the output format

%addresses, letterheads
\def\deptaddress{\hskip 3truein Department of Mathematical Sciences\par
   \noindent e-mail: {\tt fremdh}\hskip 2.9truein\today\par
   \medskip}
\long\def\univaddress#1{\hskip 3truein Department of Mathematical Sciences\par
   \hskip 3truein University of Essex\par
   \hskip 3truein Colchester CO4 3SQ\par
   \hskip 3truein England\par
   %\hskip 3truein van Vleck Hall\par
   %\hskip 3truein 480 Lincoln Drive\par
   %\hskip 3truein Madison\par
   %\hskip 3truein WI 53706-1388, U.S.A.\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
       %{\tt fremlin\@math.wisc.edu}
   \bigskip}
\long\def\homeaddress#1{\hskip 3truein 25 Ireton Road\par
   \hskip 3truein Colchester CO3 3AT\par
   \hskip 3truein England\par
   \hskip 3.2truein#1\par
   \noindent e-mail: {\tt david\@fremlin.org}
   \bigskip}
\long\def\letterhead#1#2#3{\univaddress{#1}
   {\parindent=0pt
    #2
   }
   \medskip

   \noindent Dear #3,
     }
\def\shortaddress{\noindent Departmentof Mathematical Sciences, University of Essex,
   Colchester CO4 3SQ, England\par
   \noindent e-mail: david\@fremlin.org}
\def\valediction#1#2{\bigskip
   \hskip 3truein #1,
   \vskip 1truein
   \hskip 3.3truein #2
   \bigskip}

%picture importation
   %It is clear that there is no general standard.   The syntax I
   %use when calling for a  .ps  file is
   %    \def\Caption{<...>}
   %    \picture{<filename>}{<dim>}
   %which is supposed to import  <filename>.ps  at height  <dim> ,
   %centered, and adding a one-line centered caption  <...> .
   %The basic macro listed in  \def\picture  is that which
   %used to work at Essex.   I now generally use the macro in
   % \def\picturewithpsfig , which requires a previous input
   % psfig.tex  or  psfig.sty ;
   %if you interpolate  \atUEssex  early in your file it will
   %give you the latter.
   %   I find that  psfig  is erratic in its centering, especially
   %under magnification, so I have written a routine
   % \centerpicturewithpsfig  (which may be called from
   % \centerpicture ) with the syntax
   %    \centerpicture{<filename>}{<dim>}{<dim>}
   %the first dimension being width and the second height.

   % A second package is called for by  \usegraphicx .   This requires
   % miniltx.tex  and  graphicx.sty ;  miniltx.tex  is currently (April 2016)
   % available from
   % http://mirror.ox.ac.uk/sites/ctan.org/macros/plain/graphics/miniltx.tex
   % The advantage of this is that it seems to allow manipulation of the
   % dvi>pdf stage using  dvipdfmx .

\def\picture#1#2{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at height #2
         with dvitops}
  \vbox{\vskip #2
    \special{dvitops: import #1.ps \pagewidth #2}
    \Centerline{\Caption}
    }
   \medskip
   \def\Caption{}}
%an alternative version, without caption
\def\centerpicture#1#2#3{\bigskip
  \bigskip
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with dvitops}
  \vbox{\vskip #3
    \special{dvitops: import #1.ps \pagewidth #3}
  }}
%to put pictures on same line
\def\sideshiftedpicture#1#2#3#4{\bigskip
   \bigskip
   \immediate\write0{importing diagram #1.ps at height #4
        and width #3 with dvitops}
   \vbox{\special{dvitops:  import #1.ps #3 #4}}
   \hskip #2 plus 0pt minus 0pt
   }
\def\startsideshiftedpicture{}
\newcount\scaleup
\newcount\scaledown
\scaleup=1
\scaledown=1
\def\scalesideshiftedpicture#1#2#3#4{
  \dimen4=#4
  \dimen3=#3
  \multiply\dimen4 by\scaleup
  \divide\dimen4 by \scaledown
  \multiply\dimen3 by\scaleup
  \divide\dimen3 by \scaledown
  \sideshiftedpicture{#1}{#2}{\dimen3}{\dimen4}}

\def\picturewithpsfig#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with psfig}
   \vbox{\hsize\pagewidth
   \centerline{\psfig{figure=#1.ps,height=#2}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithpsfig#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with psfig}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\psfig{file=#1.ps,width=\dimen3,height=\dimen4}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithpsfig#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with psfig}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\psfig{figure=#1.ps,height=\dimen4,width=\dimen3}}
   \psfig{figure=#1.ps,height=\dimen4,width=\dimen3}
   \hskip #2 plus 0pt minus 0pt}

\def\picturewithgraphicx#1#2{\bigskip
   \immediate\write0{importing diagram #1.ps at height #2 with graphicx}
   \vbox{\hsize\pagewidth
   \centerline{\includegraphics[height=#2]{#1.ps}}
   \medskip
   \centerline{\Caption}}
   \medskip
   \def\Caption{}}
\def\centerpicturewithgraphicx#1#2#3{
  \immediate\write0{importing diagram #1.ps at width #2
         and height #3 with graphicx}
  \dimen 3 = #2
  \dimen 4 = #3
  \dimen 2 = \pagewidth
  \advance\dimen2 by -\dimen3
  \divide\dimen2 by 2
  \discrversionB{\shrinkonestep}{}
  \discrversionH{\shrinkonestep}{}
  \discrversionF{\enlargeonestep}{}
  \discrversionG{\enlargeonestep\enlargeonestep}{}
  \dimen 6 = \dimen4
  \divide\dimen6 by 3
  \vskip\dimen6
  \ifdim\dimen2>0\hskip\dimen2\fi
     \vbox{\includegraphics[width=\dimen3,height=\dimen4]{#1.ps}}
  \discrversionG{\vskip-\dimen6}{}
  }
\def\enlargeonestep{\dimen5 = 5truein
   \dimen6 = 7truein
   \ifdim\dimen3<\dimen5
      \ifdim\dimen4<\dimen6
         \divide\dimen3 by 5
         \multiply\dimen3 by 6
         \divide\dimen4 by 5
         \multiply\dimen4 by 6
      \fi\fi}
\def\shrinkonestep{\divide\dimen3 by 6
         \multiply\dimen3 by 5
         \divide\dimen4 by 6
         \multiply\dimen4 by 5}
\def\sideshiftedpicturewithgraphicx#1#2#3#4{
   \immediate\write0{importing diagram #1.ps at height #4
      and width #3 with graphicx}
   \dimen 4 = #4
   \dimen 3 = #3
   \discrversionB{\shrinkonestep}{}
   \discrversionH{\shrinkonestep}{}
   \discrversionF{\enlargeonestep}{}
   \discrversionG{\enlargeonestep\enlargeonestep}{}
%   \vbox{\includegraphics[height=\dimen4,width=\dimen3]{#1.ps}}
   \includegraphics[height=\dimen4,width=\dimen3]{#1.ps}
   \hskip #2 plus 0pt minus 0pt}

%alternative picture importation code;  I've been told about this
  %but have not used it successfully myself
%\def\Caption{}
%\input epsf
%\epsfverbosetrue
%\def\picture#1#2{\bigskip
%    \bigskip
%    \immediate\write0{importing diagram #1.ps at height #2 with epsf}
%    \epsfysize=#2
%    \centerline{\epsffile{#1.ps}}
%    \medskip
%    \Centerline{\Caption}
%    \medskip
%    }

%in addition it may be necessary to amend the imported  .ps  file
%by replacing the   %%BoundingBox   header by  %%BoundingBox:

\def\atUEssex{
   \input psfig.sty
   \def\picture##1##2{\picturewithpsfig{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithpsfig{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithpsfig{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
   }
\def\usegraphicx{
   \input miniltx
   \def\Gin@driver{dvips.def}
   \input graphicx.sty
   \def\picture##1##2{\picturewithgraphicx{##1}{##2}}
   \def\sideshiftedpicture##1##2##3##4{
     \sideshiftedpicturewithgraphicx{##1}{##2}{##3}{##4}}
   \def\centerpicture##1##2##3{
       \centerpicturewithgraphicx{##1}{##2}{##3}}
   \def\startsideshiftedpicture{
       \sideshiftedpicture{empty}{1pt}{1pt}{1pt}}
}


%special fonts
\def\biggerscriptfonts{\scriptfont0=\eightrm
     \scriptfont1=\eighti
     \scriptfont2=\eightsy
     \scriptfont3=\eightex
     \scriptfont\bffam=\eightbf
     \scriptfont\msbfam=\eightmsb
     \scriptfont\eufmfam=\eighteufm
     }
\def\Loadcyrillic{\ifamstexsupported
      \input cyracc.def
      \font\tencyr=wncyr10
      \def\cyr{\tencyr\cyracc} \def\itcyr{\tenitcyr\cyracc}
    \else\def\cyr{\smc}\def\cdprime{u\i}\def\Cdprime{UI}\fi}
%http://wkweb3.cableinet.co.uk/linux/LDP/HOWTO/Cyrillic-HOWTO-7.html
%cyrillic.tex

\def\Loadfourteens{\iffourteensavailable
       \font\fourteenrm=cmr14
       \font\fourteenbf=cmbx14
       \font\fourteenit=cmsl14
   \else\font\fourteenrm=cmr10 scaled \magstep2
      \font\fourteenbf=cmbx10 scaled \magstep2
      \font\fourteenit=cmsl10 scaled \magstep2
   \fi}
\def\smallerfonts{\eightpoint
   \textfont\msbfam=\eightmsb
   \textfont\eufmfam=\eighteufm
   \textfont\bffam=\eightbf
   \textfont\eurmfam=\eighteurm
   \textfont\eusmfam=\eighteusm
   }
\font\twelveex=cmex10 scaled \magstep1


%Copyright declarations
\newcount\oldcopyrightdate
\newcount\newcopyrightdate
\oldcopyrightdate=0
\def\copyrightdate#1{\newcopyrightdate=#1
       \ifnum\newcopyrightdate=\oldcopyrightdate\else
       \oldcopyrightdate=#1
       \oldfootnote{}{\eightrm \copyright\ #1 D.~H.~Fremlin}
       \immediate\write0{Copyright #1}
       \fi}

%hyphenation
\hyphenation{equi-ver-id-ical}

%miscellaneous definitions
\def\Aut{\mathop{\text{Aut}}}
\def\backstep#1{\hskip -0.#1em plus 0 pt minus 0 pt}
\def\Bang{\vthsp\hbox{\bf X\hskip0.07em\llap X\hskip0.07em\llap X\rm}}
\def\BanG{\Bang{\hskip 1.111pt plus 3.333pt minus 0pt}}
%\def\BbbN{\mathchoice{\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\Bbb N$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb N$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb N$\hskip0.04em}}}
%\def\BbbQ{\mathchoice{\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\Bbb Q$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Q$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Q$\hskip0.04em}}}
%\def\BbbR{\mathchoice{\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\Bbb R$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb R$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb R$\hskip0.04em}}}
%  %do not use with $\Bbb R^X$
%\def\BbbZ{\mathchoice{\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\Bbb Z$\hskip0.02em}}
%  {\hbox{$\scriptstyle\Bbb Z$\hskip0.04em}}
%  {\hbox{$\scriptscriptstyle\Bbb Z$\hskip0.04em}}}
\def\BbbN{\openBbb{N}}
\def\BbbQ{\openBbb{Q}}
\def\BbbR{\openBbb{R}}
  %do not use with $\Bbb R^X$  do not use in $_{\Bbb R^r}$
\def\BbbZ{\openBbb{Z}}
\def\Bbbone{\hbox{1\hskip0.13em\llap{1}}}
\def\biggerint{\textfont3=\twelveex\intop\nolimits}
  % you can get this size into a displayed formula with
  %  $$\hbox{$\biggerint_{\Cal S}\pmb{u}\,d\pmb{v}$}$$
  % but it may then seem cramped
  % or try \hbox{$\biggerint_{\Cal S}\,$}
\def\biggeriint{\textfont3=\twelveex\iint}
  % remarkable fact:  this seems to infect all \int within a given
  % $..$ ; but  $\biggerint$$\int$ seems to leave the $\int$ normal
  % really  \textfont3=\twelveex  is more reliable, since  \biggerint
  % as written gets turned off rather easily (e.g. by  \comment,
  % or by an extra  {...})
\def\bover#1#2{{{#1}\over{#2}}}
\def\Bover#1#2{\backstep4\hbox{
     \biggerscriptfonts${\shortstrut#1}\over{\shortstrut#2}$}}
  % when there are superscripts in the denominator, as in \Bover1{2^m}
  % or  \Bover1{\sqrt{2\pi}^r} , these can sometimes come out with better
  % horizontal alignment if you use  \Bover1{{\sshortstrut 2}^m}  or
  % \Bover1{{\sqrt{2\pi}}^r}
\def\bsover#1#2{\textstyle\bover{#1}{#2}}
\def\bsp{/\penalty-100}
  % \penalty seems to behave oddly in a \tt string in \fullsize mode
\def\Caption{}
\def\Center#1{

   \Centerline{#1}

   \noindent}
\def\Centerline#1{\vskip 5pt
     {\def\int{\text{}$$\textfont3=\twelveex\intop\nolimits}
      \def\dashint{\text{}$$\textfont3=\twelveex
         \mskip2mu\vrule height2.65pt width4.5pt depth-2.35pt\mskip-11mu
	 \intop\nolimits\mskip-2mu}
      \def\idashint{\text{}$$\int\mskip-7mu\dashint}	 
      \def\dashiint{\text{}$$\dashint\mskip-6mu\int}	 
      \def\dashidashint{\text{}$$\dashint\mskip-6mu\dashint}	 
      \def\underlineint{
           \text{}$$\textfont3=\twelveex\underline{\intop\nolimits}}
      \def\overlineint{
           \text{}$$\textfont3=\twelveex\overline{\intop\nolimits}}
      \def\iint{\text{}$$\textfont3=\twelveex
           \intop\nolimits\hskip-0.5em\intop\nolimits}
      \def\sum{\text{}$$\sumop}
      \def\prod{\text{}$$\prodop}
      \def\bigcap{\text{}$$\bigcapop}
      \def\bigcup{\text{}$$\bigcupop}
   \centerline{#1}}\vskip 5pt}
   \mathchardef\bigcapop="1354
   \mathchardef\bigcupop="1353
   \mathchardef\prodop="1351
   \mathchardef\sumop="1350
   %in  \Centerline{$\overline{}$, $\cmmnt{}$} must use \sumop etc
\def\cf{\mathop{\text{cf}}}
\def\circumflex#1{{\accent"5E #1}}
\def\closeplus{\hbox{+}}
\long\def\cmmnt#1{\ifresultsonly\else{#1}\fi}
\def\coint#1{\left[#1\right[}
\def\curl{\mathop{\text{curl}}}
\def\dashint{\mathchoice{-\mskip-20mu\intop\nolimits}
   {{\vrule height2.75pt width4.5pt depth-2.45pt\mskip-10mu\intop\nolimits}}
      {-\mskip-16.5mu\intop\nolimits} %not yet used
      {-\mskip-16.5mu\intop\nolimits}} %not yet used
\def\idashint{\mathchoice{\int\mskip-6mu\dashint}
      {\int\mskip-5mu\dashint}
      {\int\mskip-5mu\dashint} %not yet used
      {\int\mskip-5mu\dashint}} %not yet used
\def\dashiint{\mathchoice{\dashint\mskip-10mu\int}
      {\dashint\mskip-6mu\int}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashidashint{\mathchoice{\dashint\mskip-10mu\dashint}
      {\dashint\mskip-6mu\dashint}
      {\dashint\mskip-6mu\dashint} %not yet used
      {\dashint\mskip-6mu\dashint}} %not yet used
\def\dashidashiint{\mathchoice{\dashint\mskip-8mu\dashint\mskip-10mu\int}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\dashiidashint{\mathchoice{\dashint\mskip-10mu\int\mskip-8mu\dashint}
      {\dashint\mskip-4mu\dashint\mskip-8mu\int} 
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\idashidashint{\mathchoice{\int\mskip-8mu\dashint\mskip-8mu\dashint}
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int} %not yet used
      {\dashint\mskip-6mu\int}} %not yet used
\def\diam{\mathop{\text{diam}}}
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
           \else\ifnum\stylenumber=3{\frnewpage}
           \else\ifnum\stylenumber=5{\frnewpage}
           \else\bigskip\fi\fi\fi}
\long\def\discrversionA#1#2{\ifnum\stylenumber=1#1\else#2\fi}
\long\def\discrversionB#1#2{\ifnum\stylenumber=2#1\else#2\fi}
\long\def\discrversionC#1#2{\ifnum\stylenumber=3#1\else#2\fi}
\long\def\discrversionD#1#2{\ifnum\stylenumber=4#1\else#2\fi}
\long\def\discrversionE#1#2{\ifnum\stylenumber=5#1\else#2\fi}
\long\def\discrversionF#1#2{\ifnum\stylenumber=6#1\else#2\fi}
\long\def\discrversionG#1#2{\ifnum\stylenumber=7#1\else#2\fi}
\long\def\discrversionH#1#2{\ifnum\stylenumber=8#1\else#2\fi}
           % facility for special interpolations in particular output
           % versions
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\diverg{\mathop{\text{div}}}
\def\dom{\mathop{\text{dom}}}
\def\eae{=_{\text{a.e.}}}

\def\esssup{\mathop{\text{ess sup}}}
\mathchardef\Exists="3239

\def\family#1#2#3{\langle#3\rangle_{#1\in#2}}
\def\familyi#1#2{\family{i}{#1}{#2}}
\def\familyiI#1{\familyi{I}{#1}}

\def\frfilename#1{\ifnum\stylenumber=1{\noindent filename #1}
               \else\ifnum\stylenumber=6{\noindent{\fiverm filename #1}}
               \else\ifnum\stylenumber=7{\noindent{\fiverm filename #1}}
               \else\fi\fi\fi}
\def\Footnote#1#2{\footnote"#1"{#2}}
\mathchardef\Forall="3238
\def\frakc{\frak c\phantom{.}}
\def\frnewpage{\vskip \pageheight plus 0pt minus\pageheight
           \eject
           \gdef\topparagraph{}
           \gdef\bottomparagraph{}
           }
\def\frsmallcirc{{\hskip 1pt plus 0pt minus 0pt}
     {\raise 1pt\hbox{$\smallcirc$}}{\hskip 1pt plus 0pt minus 0pt}}

\def\geae{\ge_{\text{a.e.}}}
\def\grad{\mathop{\text{grad}}}
\def\grheada{{\bf ($\pmb{\alpha}$)}}
\def\grheadb{{\bf ($\pmb{\beta}$)}}
\def\grheadc{{\bf ($\pmb{\gamma}$)}}
\def\grheadd{{\bf ($\pmb{\delta}$)}}
\def\grheade{{\bf ($\pmb{\epsilon}$)}}
\def\grheadz{{\bf ($\pmb{\zeta}$)}}
\def\grv#1{{\accent"12 #1}}

\def\hfillall{\hskip\pagewidth plus0pt minus\pagewidth}

\def\Imag{\mathop{\Cal I\text{m}}}
\long\def\inset#1{{\narrower\narrower{\vskip 1pt plus 1pt minus 0pt}
     #1
     {\vskip 1pt plus 1pt minus 0pt}}}
\def\Int#1#2{\displaystyle{\int_{#1}^{#2}}}
\def\interior{\mathop{\text{int}}}

\def\leae{\le_{\text{a.e.}}}
\long\def\leaveitout#1{}
\def\LLcorner{\mathchoice{\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height7pt width0.5pt depth0pt
       \vrule height0.5pt width5pt depth0pt}\,}
   {\,\hbox{\vrule height5pt width0.3pt depth0pt
       \vrule height0.3pt width4pt depth0pt}\,}
   {\,\hbox{\vrule height4pt width0.25pt depth0pt
       \vrule height0.25pt width3pt depth0pt}\,}
  }
\def\Ln{\mathop{\text{Ln}}}
\def\Loadfont#1{\font#1}
\def\loibr{\hbox{\hskip0.1em]\hskip0.05em}}
\def\Matrix#1{\pmatrix#1\endpmatrix}
\def\Mu{\text{M}}
\def\noamslogo{\def\amslogo{}}
\def\normalsubgroup{\vartriangleleft}
\def\Nu{\text{N}}
\def\ocint#1{\left]#1\right]}
\def\ofamily#1#2#3{\langle#3\rangle_{#1<#2}}
\let\oldcases=\cases
\let\oldfootnote=\footnote
\def\On{\mathop{\text{On}}}
\def\ooint#1{\left]#1\right[}
\def\openBbb#1{\Bbb#1\mskip1mu}
\def\pd#1#2{\bover{\partial#1}{\partial#2}}
\def\Pd#1#2{\Bover{\partial#1}{\partial#2}}
\def\pdd#1#2{{{\partial^2#1}\over{\partial#2^2}}}
\def\Pdd#1#2{\Bover{\partial^2#1}{\partial#2^2}}
\def\pde#1#2#3{{{\partial^2#1}\over{\partial#2\partial#3}}}
\def\Pde#1#2#3{\Bover{\partial^2#1}{\partial#2\partial#3}}
\def\Prf{\vthsp\hbox{\bf P\hskip0.07em\llap P\hskip0.07em\llap P\rm}}
\def\pushbottom#1{\hbox{\vrule height0pt depth#1 width0pt}}
\def\Qed{\vthsp\hbox{\bf Q\hskip0.07em\llap Q\hskip0.07em\llap Q\rm}}
\def\QeD{\Qed{\hskip 1.111pt plus 3.333pt minus 0pt}}
\def\Quer{\vthsp\hbox{\bf ?\hskip0.05em\llap ?\hskip0.05em\llap
   ?\thinspace\rm}}
\def\query{\discrversionA{\immediate\write0{query}
    \global\advance\footnotenumber by 1
    \oldfootnote{$^{\the\footnotenumber}$}{query}}{}}
\def\Real{\mathop{\Cal R\text{e}}}
\def\restr{\mathchoice
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\restriction$}\mskip 1mu}
  {\hbox{$\scriptstyle\restriction$}\mskip 1mu}
  {\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
\def\restrp{\mathchoice
  {\hbox{$\restriction$}\mskip 2mu}
  {\hbox{$\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptstyle\restriction$}\mskip 2mu}
  {\mskip-1mu\hbox{$\scriptscriptstyle\restriction$}\mskip 1mu}}
    %use in \restrp V \restrp W \restrp Y
    % \restrp\Cal P \restrp\Bbb R \restrp\frak
    % \restrp\Omega \restrp\Tau \restrp\Upsilon
    % \restrp[ \restrp\widehat{}
\def\Rho{\text{P}}
\def\roibr{\hbox{\hskip0.05em[\hskip0.1em}}
\def\RoverC{\hbox{\biggerscriptfonts${{\Bbb R}\atop{\Bbb C}}$}}
\def\sequence#1#2{\family{#1}{\Bbb N}{#2}}
\def\sequencen#1{\sequence{n}{#1}}
\def\setover#1#2{\overset{#1}\to{#2}}
\def\shortstrut{\hbox{\vrule height7.5pt depth2.5pt width0pt}}
\def\sshortstrut{\hbox{\vrule height5.8pt depth2.5pt width0pt}}
\def\smallcirc{{\scriptstyle\circ}}
\def\ssbullet{{\scriptscriptstyle\bullet}}
\def\symmdiff{\triangle}
\def\thicken#1{#1\hskip0.03em\llap #1\hskip0.03em\llap
   #1\hskip0.03em\llap #1}
\def\Tau{{\text{T}}}
\def\tbf#1{\text{\bf #1}}
\def\terror#1#2#3{#1:  `#2' for `#3'}
\def\today{\number\day.\number\month.\number\year}
\def\UEMDRR{University of Essex Mathematics Department Research Report}
\def\UEMDWP{University of Essex Mathematics Department Working Paper}
\def\versiondate#1{
    \immediate\write0{Version of #1}
    \def\vsdate{#1}
               \ifnum\stylenumber=1\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=2\hfill Version of #1
                              \medskip
               \else\ifnum\stylenumber=6\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=7\hfill{\fiverm Version of #1}
                              \medskip
               \else\ifnum\stylenumber=8\hfill Version of #1
                              \medskip
               \else\fi\fi\fi\fi\fi}
\def\versionhead{\ifnum\stylenumber=3 \hfill PRINTER'S VERSION

                        \bigskip
                   \else\ifnum\stylenumber=4 \hfill REFEREE'S VERSION

                        \bigskip
                   \else \fi\fi}
\def\versionref{\ifnum\stylenumber=1\workingref
           \else\ifnum\stylenumber=2\workingref
           \else\ifnum\stylenumber=8\workingref
           \else\printerref\fi\fi\fi}
\def\vthsp{\hskip0.001em}

%Missing AmSTeX can be coped with by using the following lines:

\ifamstexsupported\else
   \immediate\write0{AmSTeX-unsupported version of  fremtex.tex }
   \fourteensavailablefalse
   \def\@{@}
   \def\\{\cr}
   \def\AmSTeX{AmS\TeX}
   \def\Bbb#1{\underline{\hbox{\bf #1}}}
   \def\biggerscriptfonts{}
   \def\bumpeq{=_{\text{approx}}}
   \def\Cal#1{{\cal#1}}
   \def\curlyeqprec{\preceq}
   \def\eightit{\it}
   \def\eightrm{\rm}
   \def\eightsmc{\smc}
   \def\eurm#1{\underline{\cal#1}}
   \def\Footnote#1#2{\footnote#1#2}
   \def\frac#1#2{{#1}\over{#2}}
   \def\frak#1{{\hbox{\bf #1\hskip0.1em\llap #1}}}
   \def\frnewpage{\vfill\eject}
   \def\iint{\int\!\!\!\int}
   \def\iiint{\int\!\!\!\int\!\!\!\int}
   \def\iiiint{\int\!\!\!\int\!\!\!\int\!\!\!\int}
   \def\llcorner{\lfloor}
   \def\loadeurm{}
   \def\loadeusm{}
%   \def\loadeusmx#1{\def\eusm#1{\underline{\cal#1}}}
   \def\ltimes{|\hskip-0.5em\times}
   \def\eusm#1{\underline{\cal#1}}
   \def\ltimes{|\times}
   \def\Loadfont#1{}
   \def\Matrix#1{\pmatrix{#1}}
   \def\noamslogo{}
   \def\nrightarrow{\not\rightarrow}
   \def\preccurlyeq{\preceq}
   \def\query{\discrversionA{\immediate\write0{query}
                             \footnote{$^*$}{query}}{}}
   \def\restriction{|}
   \def\rtimes{\times|}
   \def\setover#1#2{#2^{#1}}
   \def\shortstrut{\strut}
   \def\smallerfonts{\eightrm}
   \def\smallfrown{\wedge}
   \def\smc{}
   \def\spcheck{\check{\phantom{x}}}
   \def\sphat{\hat{\phantom{x}}}
   \def\square{\mathchoice\sqr64\sqr64\sqr43\sqr33}
   \def\sqr#1#2{{\vcenter{\vbox{\hrule height.#2pt
          \hbox{\vrule width.#2pt height#1pt \kern#1pt
                  \vrule width.#2pt}
          \hrule height.#2pt}}}}
   \def\pmb#1{\underline{#1}}
   \def\text#1{\hbox{#1}}
   \def\therefore{\ssbullet\raise.9ex\hbox{$\ssbullet$}\ssbullet}
   \def\twelvebf{\bf}
   \def\vartriangleleft{\triangleleft}
   \def\Vdash{|\vdash}
   \def\vDash{|=}
   \fi

%End of lines for missing AmSTeX

\ifamstexsupported\input amstex
%              \documentstyle{amsppt}
               \input amsppt.sty
               \PSAMSFonts
	       \tenpoint
               %\font\eightmsb=msym8
               \font\eightmsb=msbm10
               \font\eighteufm=eufm10 at 8 pt
               \font\eighteurm=eurm8
               \font\eighteusm=eusm8
               \font\varseveneufm=eufm10 at 7 pt
               \scriptfont\eufmfam=\varseveneufm
   \def\.#1{{\accent"5F #1}}
   \def\=#1{{\accent"16 #1}}
   \gdef\nopagenumbers{\NoPageNumbers}
   \else\fi

\newcount\footmarkcount
\ifDHFamstexloaded
    \loadmsBm
    \def\eightBbb#1{{\fam\msBfam#1}}
    \loadmsFm
    \def\eightfrak#1{{\fam\msFfam#1}}
  \else
    \def\eightBbb#1{{\scriptstyle\Bbb#1}}
    \def\eightfrak#1{{\scriptstyle\frak#1}}
  \fi
\ifathome\athome\fi

%interword glue (\tenrm standard):
%  (space) 3.333 + 1.667 - 1.111
%  (,)     3.333 + 2.083 - 0.889
%  (.)     4.444 + 5.    - 0.370

%very odd fact:  \discrversionA{2.}{}  doesn't work,
%   \discrversionA{ 2.}{} does

%The false line feed introduced by my Fortran programs, leading to
%a surplus  ae  at the top of the next page, seems to be associated
%with  Fk(\032)3081  or something similar in the PostScript file.


\immediate\write0{Extract from MEASURE THEORY,
   by D.H.Fremlin, University of Essex, Colchester.
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   http://dsl.org/copyleft/dsl.txt.   This is a development version and the
   source files are not permanently archived.
   For further information contact david\@fremlin.org.}

%hyphenations
\hyphenation{Archi-me-dean}

%new style
\def\smallprintersize{\magnification=\magstep0
     \overfullrule0pt
     \def\pagewidth{468pt}
     \def\pageheight{9truein}
     \hsize\pagewidth
     \vsize\pageheight
     \nologo
     \def\copyrightdate##1{}
     \def\folio{\ifnum\pageno<0{}\else\number\pageno\fi}
     \def\mtfootline{}
     \stylenumber=10
     \pageno=-1000}

%With-proofs, results-only versions
\newif\ifwithproofs
\withproofstrue
\newif\ifresultsonly
\resultsonlyfalse
\def\noproofs{\withproofsfalse}
\def\resultsonly{\resultsonlytrue\withproofsfalse}
\long\def\proof#1{\ifwithproofs{\medskip

     \noindent{\bf proof} #1}\else\fi}
\long\def\prooflet#1{\ifwithproofs{#1}\else\fi}
%\long\def\Prooflet#1{\prooflet{\Prf\ #1 \Qed\ }}
\long\def\endnotes#1{\ifresultsonly\else{#1}\fi}
\long\def\exercises#1{\ifresultsonly\else{#1}\fi}
\long\def\references#1{\ifresultsonly\else{#1}\fi}
\long\def\dvro#1#2{\ifresultsonly{#1}\else{#2}\fi}

%Headline assignments

\def\topparagraph{}
\def\bottomparagraph{}
\def\newparagraph{}
\newif\ifpageturned

\newdimen\rhoffset
\newdimen\lhoffset
\lhoffset=0cm
\rhoffset=0cm

\newcount\volumeno
\volumeno=5
\newcount\luluvolumeno
\luluvolumeno=0

\let\plainpagebody=\pagebody
\def\pagebody{\ifodd\pageno\global\hoffset=\rhoffset
         \else\global\hoffset=\lhoffset\fi
     \plainpagebody
     \turnpage}
\def\turnpage{
     \iffirstpage\else{\baselineskip=24pt\line{\mtfootline}}\fi
%     \immediate\write0{turning page \folio,
%           top paragraph \topparagraph,   bottom \bottomparagraph,
%           new \newparagraph, pagetotal \the\pagetotal}
     \global\pageturnedtrue
     \global\firstpagefalse
     \global\let\topparagraph=\bottomparagraph
     \global\let\bottomparagraph=\newparagraph
     }

\newdimen\pagetotalplus
\newdimen\pagetotalminus

%wheader parameters:  #1 new paragraph name
 % #2#3#4  \vskip parameters if page break not forced (no. of points)
 % #5  spare height demanded if page break not to be forced
\long\def\wheader#1#2#3#4#5{\gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #5
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
%     \immediate\write0{forcing page break}
     \eject
     \else\vskip #2pt plus #3pt minus #4pt\fi
     \else\vskip #2pt plus #3pt minus #4pt\fi
  \ifpageturned\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
     \fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
%   \immediate\write0{top paragraph \topparagraph, bottom
%       \bottomparagraph, pagetotal \the\pagetotal}
  }
\def\header#1{
     \wheader{#1}{10}{4}{4}{24pt}}
\def\leader#1#2{\immediate\write0{#1 #2}
     \wheader{#1}{14}{7}{5}{24pt}
     {\bf #1 #2}}
\def\vleader#1#2#3{\immediate\write0{#2 #3}
     \wheader{#2}{14}{7}{5}{#1}
     {\bf #2 #3}}
\def\Notesheader#1{\immediate\write0{#1 Notes and comments}
     \wheader{#1 Notes}{16}{8}{5}{48pt}
     \noindent{\bf #1 Notes and comments}}
\def\vNotesheader#1#2{\immediate\write0{#2 Notes and comments}
     \wheader{#2 Notes}{16}{8}{5}{#1}
     \noindent{\bf #2 Notes and comments}}
\def\spheader#1#2#3#4#5{\header{#1#2#3#4#5}{\bf (#5)}}
\def\spstheader#1#2#3#4#5{\header{*#1#2#3#4#5}{\bf (#5)}}
\def\vspheader#1#2#3#4#5#6{
     \wheader{#2#3#4#5#6}{10}{4}{4}{#1}{\bf (#6)}}
\def\sqheader#1#2#3#4#5{\header{#1#2#3#4#5}{$\pmb{>}${\bf (#5)}}}
\def\errheader#1#2#3{
     \wheader{#3}{6}{4}{2}{24pt}
     {\bf p #1 l #2}}
\long\def\woddheader#1#2#3#4#5{
   \ifodd\pageno\wheader{#1}{#2}{#3}{#4}{#5}
   \else\vskip #2pt plus #3pt minus #4pt\fi}

\def\indexmedskip{
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by 18 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal
      \else\medskip\fi\else\medskip\fi}

\def\indexheader#1{\immediate\write0{#1}\vindexheader{#1}{24}}
\def\vindexheader#1#2{
  \gdef\newparagraph{#1}
  \pagetotalplus=\the\pagetotal
  \advance\pagetotalplus by #2 pt
  \pagetotalminus=\the\pagetotal
  \advance\pagetotalminus by -\pageshrink
  %\showthe\pagetotal \showthe\pageshrink \showthe\pagegoal
  \global\pageturnedfalse
%  \ifdim\pagetotal>\pagegoal
%      \immediate\write0{warning: pagetotal greater than pagegoal}
%      \immediate\write0{...   last paragraph reached is #1}
%      \immediate\write0{...   check for underfull vbox here}
%      \else\fi
  \ifdim\pagetotalminus<\pagegoal\ifdim\pagetotalplus>\pagegoal{\eject}
      \fi\fi
  \ifdim\pagetotal>\pagegoal\else\gdef\bottomparagraph{#1}\fi
  \ifpageturned{\global\let\topparagraph=\newparagraph
%     \immediate\write0{paragraph change on turning page}
      }\else\fi
%  \immediate\write0{top paragraph \topparagraph, bottom
%    \bottomparagraph}
  }

\def\leftheadline{\rlap{\folio}\hfill{\eightit\chaptername}\hfill
       \llap{\eightbf\topparagraph}}
\def\rightheadline{\rlap{\eightbf\bottomparagraph}\hfill
       {\eightit\headlinesectionname}\hfill\llap{\folio}}

%Footline assignments
\def\mtfootline{\ifodd\pageno\rightfootline\else\leftfootline\fi}
\def\rightfootline{\hfill\eightsmc D.H.Fremlin}
\def\leftfootline{\ifwithproofs{\eightsmc Measure Theory\hfill}
    \else{\eightsmc Measure Theory \eightit (abridged version)\hfill}
    \fi}

%Style assignments
\def\newchapter#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Chap.\ #1 {\it intro.}}
   \gdef\newparagraph{Chap.\ #1 {\it intro.}}
   \footmarkcount=0
   \gdef\sectionname{Introduction}
   \gdef\headlinesectionname{Introduction}
   \centerline{\bf Chapter #1}
   \medskip
   \centerline{\bf \chaptername}
   \medskip
   }
\def\newsection#1{\gdef\bottomparagraph{\S#1 {\it intro.}}
   \noindent{\bf #1 \sectionname}
   \smallskip
   \let\headlinesectionname=\sectionname}
\def\newvolume#1{\gdef\topparagraph{}
   \gdef\bottomparagraph{Vol.\ #1 {\it intro.}}
   \Loadfont{\twelvebf=cmbx12}
   \def\chaptername{Volume #1}
   \def\sectionname{\volumename}
   \def\headlinesectionname{\volumename}
   \centerline{\twelvebf Volume #1:  \volumename}
   \bigskip
   }

%font variations
\newif\ifvolwp
\volwpfalse
\def\tenrmstretch#1{\fontdimen3\tenrm=#1}
\font\elevenex=cmex10 at 11pt

%contents and introductions
\newif\iflargelogo
\def\vtmpb#1{}
\def\newvolume#1{}
\def\pagereference#1#2{\hfill\hbox{\ifresultsonly#2\qquad \hskip 1truein
                                                 \else#1\qquad \fi}}
\long\def\section#1#2#3#4#5#6{\ifresultsonly
    \qquad #1 #2\pagereference{#4}{#5}
  \else\vskip 2pt plus 1pt minus 0pt
    \vbox{\qquad #1 #2\pagereference{#4}{#5}\par
    \abstr{#6}
    }\vskip -1pt plus 1pt minus 0pt
  \fi}
\long\def\chapintrosection#1#2#3{\ifresultsonly
    \qquad Introduction\pagereference{#2}{#3}
  \else\vskip 1pt plus 1pt minus 0pt
    \qquad Introduction\pagereference{#2}{#3}
    \vskip 1pt plus 1pt minus 0pt
  \fi}
\def\abstr#1{\ifresultsonly{}
   \else{\advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \advance\rightskip by\parindent
     \advance\leftskip by\parindent
     \vskip1pt plus 0pt minus0pt
     \parindent=0pt
     \smallerfonts #1%
     \hbox{\vrule height0pt depth2pt width0pt}
     \vskip1pt plus 0pt minus0pt}\fi}
     
%language variations
\newif\ifUSEnglish
\USEnglishfalse
%\USEnglishtrue

%"Later editions only"
\newcount\latereditionpage
\newcount\latereditionmark
\latereditionpage=-1
\latereditionmark=0
\newcount\newpageno

%removing the \write0 instruction altogether disorganizes it!
\def\Latereditions{\newpageno=\pageno%
  \ifdim\pagetotal>\pagegoal\advance\newpageno by 1\fi%
     \divide\newpageno by 2%
  \immediate\write0{%
%     "Later editions only";
%      pagetotal \the\pagetotal, pagegoal \the\pagegoal,
%      pageno \number\pageno,\par newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage%
  }\ifnum\latereditionpage=\newpageno%
     \ifDHFamspptloaded$^{\number\latereditionmark}$%
     \else\footnote{Later editions only.}\fi%
  \else\footnote{Later editions only.}%
  \global\latereditionpage=\newpageno%
  \global\latereditionmark=\footmarkcount\fi}%
%  \immediate\write0{newpageno \number\newpageno,
%      latereditionpage \number\latereditionpage}

   
%general definitions
\def\action{\mathchoice{\lower-.3ex\hbox{$\ssbullet$}}
                       {\lower-.3ex\hbox{$\ssbullet$}}
                       {\ssbullet}
                       {\ssbullet}}  %see also \varaction
\def\add{\mathop{\text{add}}\nolimits}
\long\def\allowmorestretch#1#2{
  \ifdim\pagewidth<#1pt\else
    \fontdimen3\tenrm=2pt\fontdimen4\tenrm=1.67pt\fi
  #2
    \fontdimen3\tenrm=1.67pt\fontdimen4\tenrm=1.11pt}
\def\am{atom\discretionary{-}{}{}lessly-measurable}
\def\AmuA{\Aut_{\bar\mu}\mskip-2mu\frak A}
\def\ASint{\mathchoice{\Bbb A\mskip-5mu}
                  {\Bbb A\mskip-3mu}
                  {\Bbb A\mskip-2mu}
                  {\Bbb A\mskip-2mu}\Sint}
\def\Asisfour{$(\frak A,\bar\mu,T,\family{t}{T}{\frak A_t})$}
\def\Asissix{$(\frak A,\bar\mu,T,
   \penalty-100\family{t}{T}{\frak A_t},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Asisfourt{$(\frak A,\bar\mu,\coint{0,\infty},
   \langle\frak A_t\rangle_{t\ge 0})$}
\def\Asissixt{$(\frak A,\bar\mu,\coint{0,\infty},
   \penalty-100\langle\frak A_t\rangle_{t\ge 0},
   \penalty-100\Cal T,\family{\tau}{\Cal T}{\frak A_{\tau}})$}
\def\Atop#1#2{{\scriptstyle #1}\atop{\scriptstyle #2}}
\def\Aut{\mathop{\text{Aut}}}

\def\Bcap{\Booleanop{\cap}}
\def\Bcup{\Booleanop{\cup}}
\def\Booleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
    {\hbox{\hskip.2em\scriptfont2=\eightsy$\scriptstyle#1$\hskip.2em}}
 %   {{\mskip2mu\scriptstyle#1\mskip2mu}}  %until 15.10.17
    {{\scriptscriptstyle#1}}  %from 15.10.17
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminus{\Booleanop{\setminus}}
\def\Bsubseteq{\hskip.1em\Booleanop{\subseteq}\hskip.1em}
\def\Bsubset{\Booleanop{\subset}}
\def\Bsupseteq{\Booleanop{\supseteq}}
\def\Bsupset{\Booleanop{\supset}}
\def\Bsymmdiff{\Booleanop{\symmdiff}}
\def\Bcapshort{\Booleanopshort{\cap}}
\def\Bcupshort{\Booleanopshort{\cup}}
\def\Booleanopshort#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hskip-1pt\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\Bsetminusshort{\Booleanopshort{\setminus}}
\def\BSint{\mathchoice{\Bbb B\mskip-5mu}
                  {\Bbb B\mskip-3mu}
                  {\Bbb B\mskip-2mu}
                  {\Bbb B\mskip-2mu}\Sint}
%\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}\hskip.1em}
\def\Bsubseteqshort{\hskip.1em\Booleanopshort{\subseteq}}
\def\Bsubsetshort{\Booleanopshort{\subset}}
\def\Bsupseteqshort{\Booleanopshort{\supseteq}}
\def\Bsymmdiffshort{\Booleanopshort{\symmdiff}}

\def\closeBooleanop#1{
   \ifamstexsupported{\mathchoice
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {\hbox{\scriptfont2=\eightsy$\scriptstyle#1$}}
    {{\scriptstyle#1}}
    {{\scriptscriptstyle#1}}}
   \else{\mathchoice{{\scriptstyle#1}}{{\scriptstyle#1}}
    {{\scriptstyle#1}}{{\scriptscriptstyle#1}}}
    \fi}
\def\closeBcup{\closeBooleanop{\cup}}
\def\closeBcap{\closeBooleanop{\cap}}
\def\closeBsetminus{\closeBooleanop{\setminus}}
\def\closeBsymmdiff{\closeBooleanop{\symmdiff}}

\def\biggerHeint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 2.6pt{\sevenrm H}\hss}
   \intop\nolimits}
\def\biggerPfint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}
   \intop\nolimits\hskip 1.2pt}
\def\biggerSint{\textfont3=\twelveex
   \raise 0.6pt\hbox to 0pt{\hskip 3.6pt{\sevenrm S}\hss}
   \intop\nolimits}
\def\bu{\mathop{\text{bu}}}
\def\Bvalue#1{\mathchoice
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$[\![#1]\!]$}}
   {\hbox{$\scriptstyle[\![#1]\!]$}}
   {\hbox{$\scriptscriptstyle[\![#1]\!]$}}}
\def\cadlag{c\`adl\`ag}
\def\CalBa{\Cal{B}\eurm{a}}
   %note: neither  \cal  nor  \eusm  has lower case
\def\CalFr{\Cal F_{\text{Fr}}}
\def\callal{c\`all\`al}
\def\CalNwd{\Cal{N}\hskip-1.2pt\eurm{w}\hskip-.8pt\eurm{d}}

\def\CalRbg{\Cal{R}\mskip-1mu\eurm{b}\mskip-1mu\eurm{g}}
\def\CalSmz{\Cal{S}\mskip-2mu\eurm{mz}}
\def\CalUn{\Cal{U}\eurm{n}}
\def\capacity{\mathop{\text{cap}}\nolimits}
\def\Caratheodory{Cara\-th\'eo\-dory}
\def\Cdlg{C_{\text{dlg}}}
\def\cff{\cf\mskip1mu}
\def\checkpmb#1{\check{\pmb #1}}
\def\ci{\mathop{\text{ci}}}
\def\cl{\mathop{\text{cl}}}
\def\Clll{\hbox{$C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\clMsimp{\overline{M}_{\text{simp}}}
\def\closeuparrow{\mathchar"0222}
\def\closedownarrow{\mathchar"0223}
\def\clstar{\text{cl*}}
\def\cov{\mathop{\text{cov}}}
\def\covar#1#2{\mathchoice
  {[#1{\lower4pt\hbox{${\scriptstyle *}\atop{\vrule height 18pt depth -10pt}$}}#2]}
  {[#1{*\atop{\vrule height 8pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}
  {[#1{*\atop{\vrule height 6pt depth 0pt}}#2]}}
\def\covSh{\mathop{\text{cov}}\nolimits_{\text{Sh}}}
\def\cdownarrow{c\mskip2mu^{\downarrow}}
\def\cycle#1{(\overleftarrow{#1})}
\def\ddaggerloc{$(\ddagger)_{\text{loc}}$}
\def\ddownarrow{d\mskip2mu^{\downarrow}}
%\def\DiniD{\setover{
%  \hbox{\hskip.1em\vrule height-1pt width.5em depth1.4pt\hfil}}D}
%\def\Dinid{\hskip.1em
%  \hbox{\hskip.05em\vrule height-1.3pt width.5em depth1.7pt}\hskip-.65em D}
\def\Dinid{\hbox{$D$\hskip-8.3pt\vrule height-1pt width7pt depth1.85pt}
   \hskip1pt}
\def\DiniD{\hbox{$D$\hskip-8.3pt\vrule height9.2pt width7ptdepth-8.35pt}
   \hskip1pt}

\def\discrcenter#1#2{\ifdim\pagewidth>#1{#2}\else{

  \Centerline{#2}

  \noindent}\fi}
    %note:  in \discrcenter, make sure that #2 ends in a space
    % (if you want one to appear) and that the continuation
    % follows without
    % a space after the final } of #2, so that it will be
    % truly unindented
\def\discrpage{\ifnum\stylenumber=1{\frnewpage}  %redefined in mt.tex
        \else\ifnum\stylenumber=3{\frnewpage}
        \else\ifnum\stylenumber=5{\frnewpage}
        \else\wheader{}{18}{14}{2}{72pt}\fi\fi\fi}
\def\displaycause#1{\noalign{\noindent (#1)}}
\def\Displaycause#1{\cr\displaycause{#1}&}
\def\dotproduct{\hskip.1em\tbf{.}\hskip.1em}
\def\doubleheadrightarrow{\rightarrow\mskip-12mu\rightarrow}
\long\def\doubleinset#1{\inset{\inset{\parindent=-20pt #1}}}
\def\duparrow{d\mskip2mu^{\uparrow}}
\def\dvAformerly#1{\discrversionA{\footnote{Formerly #1.}}{}}
\def\dvAnew#1{\discrversionA{\footnote{New #1.}}{}}
\def\dvArevised#1{\discrversionA{\footnote{Revised #1.}}{}}
\def\dvrocolon#1{\dvro{#1: }{#1}}
\def\eightVVdash{\mskip5mu\vrule height 6pt depth 2pt width 0.4pt
  \mskip2mu\vrule height 6pt depth 2pt width 0.4pt
  \vrule height 2.25pt depth -1.8pt width 3.2pt\mskip2mu}
\def\embedsinto{\mskip5mu\raise1.5pt\hbox{$\subset$}\mskip-15mu
   \lower3.5pt\hbox{$\rightarrow$}\mskip5mu}
\def\energy{\mathop{\text{energy}}\nolimits}
%\eqalign:  see \S478 for breaks during \eqalign
\def\equivGT{\equiv_{\text{GT}}}
\def\equivT{\equiv_{\text{T}}}
\def\Eta{\text{H}}
\def\Expn{\Bbb E}
\long\def\Extract{\oldfootnote{}{
   \noindent{\eightit
       \ifresultsonly{Extract from {\eightsmc Measure Theory,}
          results-only version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else\ifwithproofs{Extract from {\eightsmc Measure Theory,}
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \else{Extract from {\eightsmc Measure Theory,} abridged version,
          by {\eightsmc D.H.Fremlin,}  University of Essex,
Colchester.}
       \fi\fi
   This material is copyright.   It is issued under the terms of the
   Design Science License as published in
   {\eighttt http://dsl.org/copyleft/dsl.txt}.
   This is a development version and the
   source files are not permanently archived, but current versions are
normally
   accessible through
   {\eighttt https://www1.essex.ac.uk/maths/people/fremlin/mt.htm}.
   For further information contact {\eighttt david\@fremlin.org}.}
   }}
\def\Fn{\mathop{\text{Fn}}\nolimits}
\def\FN{\mathop{\text{FN}}\nolimits}
\def\formerly#1{\footnote{Formerly #1.}}
  %use only once in a chapter, never in same volume as reference
\def\fraction#1{\mathchoice{\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$<$}#1\hbox{$>$}}
     {\hbox{$\scriptstyle<$}#1\hbox{$\scriptstyle>$}}
     {\hbox{$\scriptscriptstyle<$}#1\hbox{$\scriptscriptstyle>$}}}
\def\frakmctbl{\frak m_{\text{countable}}}
\def\frmedcirc{{\raise 2.5pt\hbox{$_{\bigcirc}$}}}
\def\grad{\mathop{\text{grad}}\nolimits}
\def\hatpmb#1{\hat{\pmb{#1}}}
\def\hattilde#1{\skew2\hat{\tilde#1}}
\def\hattildepmb#1{\skew1\hat{\tildepmb#1}}
\def\hcov{\mathop{\text{hcov}}\nolimits}
\def\Heint{\mathchoice{\hbox to 0pt{\hskip 3.1pt H\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm H}\hss}}
     \intop\nolimits}
\def\high#1{\lceil#1\rceil}
\def\highoverline#1{\overline{\vrule height6pt width0pt depth 0pt#1}}
\def\Hint#1{({\it Hint\/}: #1)}
\def\hL{\mathop{\text{hL}}}
\def\Hmp{\text{Hom}_{\bar\mu,\pi}}
\def\hp{\mathop{\text{hp}}\nolimits}
\def\hwtr{hereditarily weakly $\theta$-{\vthsp}refinable}
\def\iiv{ii_{\pmb{v}}}
\def\iivu{\iiv(\pmb{u})}
\def\imp{inverse-{\vthsp}measure-{\vthsp}preserving}
\def\innerprod#1#2{(#1|#2)}
\def\intstar{\text{int*}}
\def\CalIS{\Cal I(\Cal S)}
\def\jump{\mathop{\text{jump}}\nolimits}
\def\Jump{\mathop{\text{Jump}}\nolimits}
\def\Krein{Kre\v{\i}n}
\def\Lambdapv{\Lambda_{\text{pv}}}
\def\leRB{\le_{\text{RB}}}
\def\leRK{\le_{\text{RK}}}
\def\link{\mathord{\text{link}}}
\def\locmafp{\widehat{\otimes}_{\text{loc}}}
\def\Locmafp{\widehat{\bigotimes}^{\text{loc}}}
\def\low#1{\lfloor#1\rfloor}
\def\lti{left-\vthsp translation-\vthsp invariant}
\def\magnitude{\mathop{\text{mag}}}
\def\Mah{\mathop{\text{Mah}}}
\def\Mahcr{\Mah_{\text{cr}}}
\def\MahcrR{\Mah_{\text{crR}}}
\def\MahqR{\Mah_{\text{qR}}}
\def\MahR{\Mah_{\text{R}}}
\def\MahspcrR{\Mah_{\text{spcrR}}}
\def\Malm{M_{\text{alm}}}
\def\Marik{Ma\v r\'\i k} 
\def\Mbv{M_{\text{bv}}}
\def\Mcdlg{M_{\text{cdlg}}}
\def\Mdm{M_{\text{D-M}}}
\def\med{\mathop{\text{med}}\nolimits}
\def\Mfa{M_{\text{fa}}}
\def\Mhsr{\mathop{\text{Mhsr}}\nolimits}
\def\Mjf{\mathop{M_{\text{j-f}}}\nolimits}
\def\Mlbv{M_{\text{lbv}}}
\def\Mlob{M_{\text{lo-b}}}
\def\Mlns{M_{\text{ln-s}}}
\def\Mmo{M_{\text{mo}}}
\def\Mlmo{M_{\text{lmo}}}
\def\Mns{M_{\text{n-s}}}
\def\Mnsu{M_{\text{n-s}}^{\uparrow}}
%\def\Mnsu{M_{\text{n-s}\uparrow}}
\def\MMah{M_{\text{Mah}}}
\def\Mob{M_{\text{o-b}}}
\def\Mpob{M_{\text{po-b}}}
\def\Mpv{M_{\text{pv}}}
\def\MSi{M_{\text{S-i}}}
\def\Msimp{M_{\text{simp}}}
\def\Mth{Ma\-ha\-ram-{\vthsp}type-{\vthsp}homogeneous}
\def\Mvlm{M_{\text{vlm}}}
\def\negquad{\hskip-1em\relax}
\def\NN{\BbbN^{\Bbb N}}
\def\non{\mathop{\text{non}}}
\def\notBooleanop#1{
   \ifamstexsupported{\mathchoice
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
{\hbox{\hskip.2em$\not\scriptfont2=\eightsy\scriptstyle#1$\hskip.2em}}
    {{\not\scriptstyle#1}}
    {{\not\scriptscriptstyle#1}}}
   \else{\mathchoice{{\not\scriptstyle#1}}{{\not\scriptstyle#1}}
    {{\not\scriptstyle#1}}{{\not\scriptscriptstyle#1}}}
    \fi}
\def\notBsubseteq{\notBooleanop{\subseteq}}
\def\notBsupseteq{\notBooleanop{\supseteq}}
\def\notVVdash{\not\mskip4mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip-2mu}
\def\nuprime{\nu^{\mskip0.7mu\prime}} %better  \nu'  in superscripts
\def\nw{\mathop{\text{nw}}}
\def\Osclln{\mathop{\text{Osclln}}\nolimits}
\def\otp{\mathop{\text{otp}}}
\def\partstar{\mathchoice{\partial\hbox{*}}
     {\partial\hbox{*}}
     {\partial\hbox{\sevenrm *}}
     {\partial\hbox{\fiverm *}}}
\def\per{\mathop{\text{per}}}
\def\Pfint{\mathchoice{\hbox to 0pt{\hskip 3.1pt Pf\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.3pt{\sevenrm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.0pt{\fiverm Pf}\hss}}
     \intop\nolimits
     \mathchoice{\hskip 2pt}{\hskip 1.2pt}{\hskip 1pt}{\hskip 1pt}}
\def\Pou{\mathop{\text{Pou}}\nolimits}
\def\prforc{\preccurlyeq_{\text{forcing}}}
\def\prGT{\preccurlyeq_{\text{GT}}}
\def\prT{\preccurlyeq_{\text{T}}}
\def\pssqa{power set $\sigma$-quotient algebra}
\def\qm{quasi-{\vthsp}measurable}
\def\Reverse#1{\setover
   {\lower0.4ex\hbox{$\scriptscriptstyle\leftrightarrow$}}{#1}}
\def\rhoKR{\rho_{\text{KR}}} 
\def\rhotv{\rho_{\text{tv}}} 
\def\Rint{\mathchoice{\hbox to 0pt{\hskip 3.1pt R\hss}}
     {\raise 0.6pt\hbox to 0pt{\hskip 1.9pt{\sevenrm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm R}\hss}}
     \intop\nolimits}
\def\RO{\text{RO}}
\def\rti{right-{\vthsp}translation-{\vthsp}invariant}
\def\rvm{real-{\vthsp}valued-{\vthsp}measurable}
\def\sat{\mathop{\text{sat}}}
\def\sfamilysS#1{\family{\sigma}{\Cal S}{#1_{\sigma}}}
\def\sfcc{$\sigma$-finite-cc}
\def\sgn{\mathop{\text{sgn}}}
\def\shr{\mathop{\text{shr}}\nolimits}
\def\Sii{\text{S}ii}
\def\Siiv{\Sii_{\pmb v}}
\def\Sint{\mathchoice{\hbox to 0pt{\hskip 4.1pt\ninerm S\hss}} 
     {\raise 0.6pt\hbox to 0pt{\hskip 3.0pt{\sevenrm S}\hss}}
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     {\raise 0.4pt\hbox to 0pt{\hskip 1.5pt{\fiverm S}\hss}} %unchecked
     \intop\nolimits}
\def\Sinteightpoint{\raise 0.6pt\hbox to 0pt{\hskip 2.3pt{\fiverm S}\hss}
     \intop\nolimits}
\def\SL{\text{SL}}
\def\sm{semi-martingale} %hyphen to remind me that it's the "virtual" kind
\def\Sous{\Cal S\text{ous}}  %used in vol 6
\def\spread{\mathop{\text{spr}}}
\def\ssplrarrow{^{\scriptscriptstyle\leftrightarrow}}
\def\ssptilde{{\raise1.2ex\hbox{$\scriptscriptstyle\sim$}}}
%\def\sspvec{{\raise1.2ex\hbox{$\scriptscriptstyle\rightarrow$}}}
\def\sspvec{\mskip3mu\vec{\null}\mskip5mu}
\def\Sti{\mathop{\text{Sti}}\nolimits}
\def\strbar{^{\scriptscriptstyle\pmb{-}}}
\def\strplus{^{\scriptscriptstyle\pmb{+}}}
\def\strprime{^{\mathchoice{\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1.5mu\pmb{\prime}}
                  {\mskip1mu\pmb{\prime}}}}
%\def\strprime{\,\pmb{\prime}}
\def\supp{\mathop{\text{supp}}}
\def\tausm{\tau_{\sigma\text{-m}}}
\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_<$}
   \phantom{\tbf 1^{(#1)}}}
%\def\tbfoneminus#1{\rlap{$\tbf 1^{(#1)}$}\rlap{\hskip7pt$_-$}
%   \phantom{\tbf 1^{(#1)}}}
\def\tensorhat{\widehat{\otimes}}
\def\Tensorhat{\widehat{\bigotimes}}
\def\tensorhatsigma{\widehat{\otimes}_{\sigma}}
\def\tildeClll{\hbox{$\tilde C$
  \hskip-4.5pt\raise4pt\hbox{\char'23
  \hskip-1.6pt\vrule height6.5pt width.5pt depth0pt
  \hskip.7pt\vrule height6.5pt width.5pt depth0pt
  \hskip-2.5pt\lower5.4pt\hbox{\char'23}}}}
\def\tildehat#1{\skew2\tilde{\hat#1}}
\def\tildehatpmb#1{\tilde{\hatpmb#1}}
\def\tildepmb#1{\tilde{\pmb{#1}}}
\def\Tr{\mathop{\text{Tr}}}
\def\trs{^{\top}\hskip-1pt}
\def\triplepc#1#2#3{(#1,#2,\hbox{$<$}#3)}
\def\TSi{\frak T_{\text{S-i}}}
\def\undphi{\underline{\phi}{}}
\def\undpsi{\underline{\psi}{}}
\def\undtheta{\underline{\theta}{}}
\def\updown{\mathchoice
  {\hbox{\hskip-1.5pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-3.3pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %OK for subscript, fails for superscript
    \hskip-1.7pt\vrule height6.5pt width.7pt depth0pt
    \hskip.9pt\vrule height6.5pt width.7pt depth0pt
    \hskip-2.5pt\lower5.4pt\hbox{\char'23}\hskip-1.4pt}}
  {\hbox{\hskip-2pt\char'23  %not tested
    \hskip-1.7pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip.9pt\vrule height6.5pt width.7pt depth1.3pt
    \hskip-2.5pt\lower6.7pt\hbox{\char'23}\hskip-1.4pt}}}
\def\updownarrows{\uparrow\mskip-4mu\downarrow}
\def\updowncgt{$\updown$-convergent}
\def\upr{\mathop{\text{upr}}\nolimits}
\def\ureg{\mathop{\text{ureg}}}

\def\Var{\mathop{\text{Var}}\nolimits}

\def\varaction{\mathchoice{\lower-.2ex\hbox{$\smallcirc$}}
                       {\lower-.2ex\hbox{$\smallcirc$}}
                       {\ssbullet}
                       {\ssbullet}}
\def\varcheck#1{{\setover
       {\lower 0.4ex\hbox{$\scriptscriptstyle\vee$}}{#1}}}
\def\varcheckf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\vee$}}{f}}}
\def\varchecki{\mskip-3mu\text{\it{\accent"14{\hbox{\hskip-3.5pt\i}}}}\mskip2mu}
\def\varspcheck{^{\scriptscriptstyle\vee}}
%\def\varspcheck{{\setover{\scriptscriptstyle\vee}{\phantom{g}}}}
%\def\varspcheckf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\vee$}}{\phantom{f}}}}
\def\varhat#1{{\setover{\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{#1}}}
\def\varhatf{{\setover{\,\lower 0.4ex
              \hbox{$\scriptscriptstyle\wedge$}}{f}}}
\def\varinnerprod#1#2{#1\dotproduct#2}
\def\varsphat{^{\scriptscriptstyle\wedge}}
% or use  \hbox{\^{\null}}
%\def\varsphat{{\setover{\scriptscriptstyle\wedge}{\phantom{g}}}}
%\def\varsphatf{{\setover{\lower 0.4ex
%              \hbox{$\scriptscriptstyle\wedge$}}{\phantom{f}}}}
\def\varBspcheck{\mathchoice
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2pt
  {\fourteenrm\hbox{\lower2pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}
  {\hskip-2.5pt
  {\fourteenrm\hbox{\lower4pt\hbox{\v{\null}}}}\hskip-1pt}}
\def\varCspcheck{\mathchoice
  {\hskip-1pt\check{\phantom x}}   %not tested
  {\hskip-1pt\check{\phantom x}}
  {\hskip-1.2pt\check{\phantom x}}
  {\hskip-1.5pt\check{\phantom x}}}  %not tested
\def\varCsphat{\mathchoice
  {\hskip-1pt\hat{\phantom x}}   %not tested
  {\hskip-1pt\hat{\phantom x}}
  {\hskip-1.2pt\hat{\phantom x}}  %not tested
  {\hskip-1.5pt\hat{\phantom x}}}  %not tested
\def\VeqL{$V\hbox{$=$}L$}
\def\VVdash{\mskip5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \mskip2.5mu\vrule height 7.5pt depth 2.5pt width 0.5pt
  \vrule height 2.75pt depth -2.25pt width 4pt\mskip2mu}
\def\VVdP{\VVdash_{\Bbb P}}
\def\VVdPk{\VVdash_{\Bbb P_{\kappa}}}
\def\wdistr{\mathop{\text{wdistr}}}
\def\weight{\mathop{\text{wt}}}
\def\w1lim{\mathop{\text{w1lim}}}
\def\wsid{weakly $(\sigma,\infty)$-{\vthsp}distributive}
\def\1lim{\mathop{1\mskip-1.5mu\text{lim}}}
\def\2vm{two-{\vthsp}valued-{\vthsp}measurable}

%Cara\-th\'e\-o\-dory

\def\chaptername{}
\def\sectionname{}
\def\headlinesectionname{}

\loadeusm\loadeurm
\Loadfourteens
\Loadfont{\ninerm=cmr9}

%note:  $..\cmmnt{=..}..$  is better as  \cmmnt{\mskip5mu
 
  
%----------------------------------------------------------%  
  
\smallprint  
%\USsmallprint  
%\noproofs  
%\resultsonly  
%\atUEssex  
\usegraphicx
  
\Extract  
  
%----------------------------------------------------------%  

\frfilename{mt1.tex}
\versiondate{30.9.02}
\copyrightdate{1994}

\def\volumename{The irreducible minimum}

\newvolume{1}


In this introductory volume I set out, at a level which I hope will be
suitable for students with no prior knowledge of the Lebesgue (or even
Riemann) integral
and with only a basic (but thorough) preparation in the techniques of
$\epsilon$-$\delta$ analysis, the theory of measure and integration up
to the convergence theorems (\S123).   I add a third chapter (Chapter
13) of miscellaneous additional results, mostly chosen as being
relatively elementary material necessary for topics treated in Volume 2
which does not have a natural place there.

The title of this volume is a little more emphatic than I should care to
try to justify {\it au pied de la lettre}.   I would certainly
characterize the construction of Lebesgue measure on $\Bbb R$ (\S114),
the definition of the integral on an abstract measure space (\S122)
and the convergence theorems (\S123) as indispensable.   But a teacher
who wishes to press on to further topics will find that much of Chapter
13 can be set aside for a while.   I say `teacher' rather than
`student' here, because if you are studying on your own I think you
should aim to go slower than the text requires rather than faster;  in
my view, these ideas are genuinely difficult, and I think you should
take the time to get as much practice at relatively elementary levels as
you can.

Perhaps this is a suitable moment at which to set down some general
thoughts on the teaching of measure theory.   I have been teaching
analysis for over thirty years now, and one of the few constants over
that period has been the feeling, almost universal among teachers of
analysis, that we are not serving most of our students well.   We have
all encountered students who are not stupid -- who are indeed quite good
at mathematics -- but who seem to have a disproportionate difficulty
with rigorous analysis.   They are so exhausted and demoralised by the
technical problems that they cannot make sense or use even of the
knowledge they achieve.   The natural reaction to this is to try
to make courses shorter and easier.   But I think that this makes it
even more likely that at the end of the semester your students will be
stranded in thorn-bushes half way up the mountain.   Specifically, with
Lebesgue measure, you are in danger of spending twenty hours teaching
them how to integrate the indicator function of the rationals.
This is not what the subject is for.   Lebesgue's own presentations of
the subject ({\smc Lebesgue 1904}, {\smc Lebesgue 1918}) emphasize the
convergence theorems and the Fundamental Theorem of Calculus.   I
have put the former in Volume 1 and the latter in Volume 2, but it does
seem to me that unless your students themselves want to know when one
can expect to be able to interchange a limit and an integral, or which
functions are indefinite integrals, or what the completions of
$C([0,1])$ under the norms $\|\,\|_1$,  $\|\,\|_2$  look like, then it
is going to be very difficult for them to make anything of this
material;  and if you really cannot reach the point of explaining at
least a couple
of these matters in terms which they can appreciate, then it may not be
worth starting.  I would myself choose rather to omit a good many proofs
than to come to the theorems for which the subject was created so late
in the course that two thirds of my class have already given up before
they are covered.

Of course I and others have followed that road too, with no better
results (though usually with happier students) than we obtain by dotting
every {\it i} and crossing every {\it t} in the proofs.   Nearly every
time I am
consulted by a non-specialist who wants to be told a theorem which will
solve his problem, I am reminded that pure mathematics in general, and
analysis in particular, does not lie in the {\it theorems} but in the
{\it proofs}.   In so far as I have been successful in answering such
questions, it has usually been by making a trifling adjustment to a
standard argument to produce a non-standard theorem.
The ideas are in the details.   You have not understood \Caratheodory's
construction (\S113) until you can, at the very least, reliably
reproduce the argument which shows that it works.   In the end, there is
no alternative to going over every step of the ground, and while I
have occasionally been ruthless in cutting out topics which seem to me
to be marginal, I have
tried to make sure -- at the expense, frequently, of pedantry -- that
every necessary idea is signalled.

Faced, therefore, with any particular class, I believe that a teacher
must compromise between scope and completeness.   Exactly which
compromises are most appropriate will depend on factors
which it would be a waste of time for me to guess at.   This volume is
supposed
to be a possible text on which to base a course;  but I hope that no
lecturer will set her class to read it at so many pages a week.   My
primary aim is to provide a concise and coherent basis on which to erect
the structure of the later volumes.   This involves me in pursuing, at
more than one point, approaches which take slightly more difficult paths
for the sake of developing a more refined technique.   (Perhaps the most
salient of these is my insistence that an integrable function need not
be defined everywhere on the underlying measure space;  see
\S\S121-122.)
It is the responsibility of the individual teacher to decide for herself
whether such refinements are appropriate to the needs of her students,
and, if not, to show them what translations are needed.

The above paragraphs are directed at teachers who are, supposedly,
competent in the subject -- certainly past the level treated in this
volume -- and who have access to some of the many excellent books
already available, so that if they take the trouble to think out their
aims, they should be able to choose which elements of my presentation
are suitable.   But I must also consider the position of a student who
is setting out to learn this material on his own.   I trust that you
have understood from what I have already written that you should not be
afraid to look ahead.   You could, indeed, do worse than go to Volume 2,
and take one of the
wonderful theorems there -- the Fundamental Theorem of Calculus
(\S222), for instance, or, if you are very ambitious, the strong
law of large numbers (\S273) -- and use the index and the
cross-references to try to extract a proof from first principles.   If
you are successful you
will have every right to congratulate yourself.   In the periods in
which success seems elusive, however, you should be working
systematically through the `basic exercises' in the sections which
seem to be
relevant;  and if all else fails, start again at the beginning.
Mathematics is a difficult subject, that is why it is worth doing, and
almost every section here contains some essential idea which you could
not expect to find alone.

\frnewpage
\frfilename{mt11.tex}
\versiondate{4.1.04}
\copyrightdate{1994}

\def\chaptername{Measure spaces}
\def\sectionname{Introduction}

\newchapter{11}

In this chapter I set out the fundamental concept of `measure space',
that is, a set in which some (not, as a rule, all) subsets may be
assigned a `measure', which you may wish to interpret as area, or
mass, or volume, or thermal capacity, or indeed almost anything which
you would expect to be additive -- I mean, that the measure of the union
of two disjoint sets should be the sum of their measures.   The actual
definition (in 112A) is not obvious, and depends essentially on certain
technical features which make a preparatory section (\S111) advisable.
Furthermore, even with the definition well in hand, the original and
most important examples of measures, Lebesgue measure on Euclidean
space, remain elusive.   I therefore devote a section (\S113) to a
method of constructing measures, before turning to the details of the
arguments needed for Lebesgue measure in \S\S114-115.   Thus the
structure of the chapter is three sections of general theory followed by
two (which are closely similar) on particular examples.   I should say
that the general theory is essentially easier;  
but it does rely on facility with certain manipulations of families of 
sets which may be new to you.

At some point I ought to comment on my arrangement of the material, and
it may be helpful if I do so before you start work on this chapter.
One of the many fundamental questions which any author on the subject
must decide, is whether to begin with `general' measure theory or
with `Lebesgue' measure and integration.   The point is that Lebesgue
measure is rather more than just the most important example of a measure
space.   It is so close to the heart of the subject that the great
majority of the ideas of the elementary theory can be fully realised
in theorems about Lebesgue measure.   Looking ahead to Volume 2, I find
that, with the exception of Chapter 21 -- which is specifically devoted
to extending your ideas of what measure spaces can be -- only
Chapter 27 and the second half of Chapter 25 really need the general
theory to make sense, while
Chapters 22, 26 and 28 are specifically about Lebesgue measure.   Volume
3 is another matter, but even there more than half the mathematical
content can be expressed in terms of Lebesgue measure.   If you
take the view, as I certainly do when it suits my argument, that the
business of pure mathematics is to express and extend the logical
capacity of the human mind, and that the actual theorems we work through
are merely vehicles for the ideas, then you can correctly point out that
all the really important things in the present volume can be done
without going to the trouble of formulating a general theory of abstract
measure spaces;  and that by studying the relatively concrete example of
Lebesgue measure on
$r$-dimensional Euclidean space you can avoid a variety of irrelevant
distractions.

If you are quite sure, as a teacher, that none of your pupils will wish
to go beyond the elementary theory, there is something to be said for
this view.   I believe, however, that it becomes untenable if you wish
to prepare any of your students for more advanced ideas.   The
difficulty is that, with the best will in the world, anyone who has
worked through the full theory of Lebesgue measure, and then comes to
the theory of abstract measure spaces, is likely to go through it too
fast, and at the end find himself uncertain about just which ninety per
cent of the facts he knows are generally applicable.   I believe it is
safer to keep the special properties of Lebesgue measure clearly
labelled as such from the beginning.

It is of course the besetting sin of mathematics teachers at this level,
to teach a class of twenty in a manner appropriate to perhaps two of
them.   But in the present case my own judgement is that very few
students who are ready for the course at all will have any difficulty
with the extra level of abstraction involved in `Let $(X,\Sigma,\mu)$
be a measure space, $\ldots$'.   I do assume knowledge of elementary
linear algebra, and the grammar, at least, of arbitrary measure spaces
is no worse than the grammar of arbitrary linear spaces.   Moreover, the
Lebesgue theory already involves statements of the form `if $E$ is a
Lebesgue measurable set, $\ldots$', and in my experience students who
can cope with quantification over subsets of the reals are not deterred
by quantification over sets of sets (which anyway is necessary for any
elementary description of the $\sigma$-algebra of Borel sets).   
So I believe
that here, at least, the extra generality of the `professional'
approach is not an obstacle to the amateur.

I have written all this here, rather than later in the chapter, because
I do wish to give you the choice.   And if your choice is to learn the
Lebesgue theory first, and leave the general theory to later, this is
how to do it.    You should read

\qquad paragraphs 114A-114C %114A 114B 114C

\qquad 114D, with 113A-113B and 112Ba, 112Bc

\qquad  114E, with 113C-113D, 111A, 112A, 112Bb

\qquad  114F

\qquad  114G, with 111G and 111C-111F, %111C 111D 111E 111F

\noindent and then
continue with Chapter 12.   At some point, of course, you should look
at the exercises for \S\S112-113;  but, as in Chapters 12-13, you will
do so by translating `Let $(X,\Sigma,\mu)$ be a measure space' into
`Let $\mu$ be Lebesgue measure on $\Bbb R$, and $\Sigma$ the 
$\sigma$-algebra of Lebesgue measurable sets'.
Similarly, when you look at 111X-111Y, you will take $\Sigma$ to be {\it
either} the $\sigma$-algebra of Lebesgue measurable sets {\it or} the
$\sigma$-algebra of Borel subsets of $\Bbb R$.

\discrpage

\frfilename{mt111.tex}
\versiondate{26.1.05}
\copyrightdate{1994}

\def\chaptername{Measure spaces}
\def\sectionname{$\sigma$-algebras}

\newsection{111}

In the introduction to this chapter I remarked that a measure space is
`a set in which some (not, as a rule, all) subsets may be assigned a
measure'.   All ordinary concepts of `length' or `area' or `volume'
apply only to reasonably regular sets.   Modern measure theory
is remarkably powerful in that an extraordinary variety of sets are
regular enough to be measured;  but we must still expect some
limitation, and when studying any measure a proper understanding of the
class of sets which it measures will be central to our work.   The basic
definition here is that of `$\sigma$-algebra of sets';  all measures
in the standard theory are defined on such collections.   I
therefore begin with a statement of the definition, and a brief
discussion of the properties, of these classes.


\leader{111A}{Definition} Let $X$ be a set.   A {\bf $\sigma$-algebra of
subsets of $X$}\cmmnt{ (sometimes called a {\bf $\sigma$-field})} is a
family $\Sigma$ of subsets of $X$ such that

\qquad(i) $\emptyset\in\Sigma$;

\qquad(ii) for every $E\in\Sigma$, its complement $X\setminus E$ in $X$
belongs to $\Sigma$;

\qquad(iii) for every sequence $\langle E_n\rangle_{n\in\Bbb N}$ in
$\Sigma$, its union $\bigcup_{n\in\Bbb N}E_n$ belongs to $\Sigma$.


\cmmnt{
\leader{111B}{Remarks (a)} Almost any new subject in pure
mathematics is
likely to begin with definitions.   At this point there is no substitute
for rote learning.   These definitions encapsulate years, sometimes
centuries, of thought by many people;  you cannot expect that they will
always correspond to familiar ideas.


\header{111Bb}{\bf (b)} Nevertheless, you should always seek immediately
to find ways
of making new definitions more concrete by finding examples within your
previous mathematical experience.   In the case of `$\sigma$-algebra',
the really important examples, to be described below, are going to be
essentially new -- supposing, that is, that you need to read this
chapter at all.   However, two examples should be immediately accessible
to you, and you should bear these in mind henceforth:

\qquad(i) for any $X$, $\Sigma=\{\emptyset,X\}$ is a $\sigma$-algebra of
subsets of $X$;

\qquad(ii) for any $X$, $\Cal PX$, the set of all subsets of $X$, is a
$\sigma$-algebra of subsets of $X$.

\noindent These are of course the smallest and largest $\sigma$-algebras
of subsets of $X$, and while we shall spend little time with them, both
are in
fact significant.


\header{111Bc}{\bf *(c)} The phrase {\bf measurable space} is often
used to mean a pair $(X,\Sigma)$, where $X$ is a set and $\Sigma$ is a
$\sigma$-algebra of subsets of $X$;  but I myself prefer to avoid this
terminology, unless greatly pressed for time, as in fact many of the
most interesting examples of such
objects have no useful measures associated with them.
}%end of comment

\cmmnt{
\leader{111C}{Infinite unions and intersections} If you have not
seen infinite unions before, it is worth
pausing over the formula $\bigcup_{n\in\Bbb N}E_n$.   This is the set of
points belonging to one or more of the sets $E_n$;  we may write it as

$$\eqalign{\bigcup_{n\in\Bbb N}E_n
&=\{x:\exists\enskip n\in\Bbb N,\,x\in E_n\}\cr
&=E_0\cup E_1\cup E_2\cup\ldots.\cr}$$

\noindent (I write $\Bbb N$ for the set of natural numbers
$\{0,1,2,3,\ldots\}$.)   In the same way,

$$\eqalign{\bigcap_{n\in\Bbb N}E_n
&=\{x:x\in E_n\Forall n\in\Bbb N\}\cr
&=E_0\cap E_1\cap E_2\cap\ldots.\cr}$$

\noindent It is characteristic of the elementary theory of measure
spaces that it demands greater facility with the set-operations $\cup$,
$\cap$, $\setminus$ (`set difference':  $E\setminus F=\{x:x\in
E,\,x\notin F\}$), $\symmdiff$ (`symmetric difference':  $E\symmdiff
F=(E\setminus F)\cup(F\setminus E)=(E\cup F)\setminus(E\cap F)$) than
you have probably needed before, with the added complication of infinite
unions and intersections.   I strongly advise spending at least a little
time with Exercise 111Xa at some point.
}%end of comment

\leader{111D}{Elementary properties of $\sigma$-algebras} If $\Sigma$ is
a $\sigma$-algebra of subsets of $X$, then it
has the following properties.

\header{111Da}{\bf (a)} $E\cup F\in\Sigma$ for all $E$, $F\in\Sigma$.
\prooflet{\Prf\ For
if $E$, $F\in\Sigma$, set $E_0=E$, $E_n=F$ for $n\ge 1$;  then
$\langle E_n\rangle_{n\in\Bbb N}$ is a sequence in $\Sigma$ and $E\cup
F=\bigcup_{n\in\Bbb N}E_n\in\Sigma$.\ \Qed}


\header{111Db}{\bf (b)} $E\cap F\in\Sigma$ for all $E$, $F\in\Sigma$.
\prooflet{\Prf\ By
(ii) of the definition in 111A, $X\setminus E$ and $X\setminus
F\in\Sigma$;  by (a) of this paragraph, $(X\setminus E)\cup(X\setminus
F)\in\Sigma$;  by 111A(ii) again, $X\setminus((X\setminus
E)\cup(X\setminus F))\in\Sigma$;  but this is just $E\cap F$.\ \Qed}


\header{111Dc}{\bf (c)} $E\setminus F\in\Sigma$ for all $E$,
$F\in\Sigma$.
\prooflet{\Prf\ $E\setminus F=E\cap(X\setminus F)$.\ \Qed}


\header{111Dd}{\bf (d)} Now suppose that
$\langle E_n\rangle_{n\in\Bbb N}$ is a
sequence in $\Sigma$, and consider

$$\eqalign{\bigcap_{n\in\Bbb N}E_n
&=\{x:x\in E_n\Forall n\in\Bbb N\}\cr
&=E_0\cap E_1\cap E_2\cap\ldots\cr
&=X\setminus\bigcup_{n\in\Bbb N}(X\setminus E_n);\cr}$$

\noindent this also belongs to $\Sigma$.


\cmmnt{
\leader{111E}{More on infinite unions and
intersections (a)} So far I
have considered infinite unions and intersections only in the context of
sequences $\langle
E_n\rangle_{n\in\Bbb N}$ indexed by the set $\Bbb N$ of natural numbers
itself.   Many others will arise more or less naturally in the pages
ahead.   Consider, for instance, sets of the form

\Centerline{$\bigcup_{n\ge 4}E_n=E_4\cup E_5\cup E_6\cup\ldots$,}

\Centerline{$\bigcup_{n\in\Bbb Z}E_n
=\{x:\exists\enskip n\in\Bbb Z,\,x\in E_n\}
=\ldots\cup E_{-2}\cup E_{-1}\cup E_0\cup E_1\cup E_2\cup\ldots$,}

\Centerline{$\bigcup_{q\in\Bbb Q}E_q
=\{x:\exists\enskip q\in\Bbb Q,\,x\in E_q\}$,}

\noindent where I write $\Bbb Z$ for the set of all integers and
$\Bbb Q$
for the set of rational numbers.   If every $E_n$, $E_q$ belongs to a
$\sigma$-algebra $\Sigma$, so will these unions.   On the other hand,

\Centerline{$\bigcup_{t\in[0,1]}E_t
=\{x:\exists\enskip t\in [0,1],\,x\in E_t\}$}

\noindent may fail to belong to a $\sigma$-algebra containing every
$E_t$, and it is of the greatest importance to develop an intuition for
those index sets, like $\Bbb N$, $\Bbb Z$ and $\Bbb Q$, which are `safe'
in this context, and those which are not.

\spheader 111Eb I rather hope that you
have seen enough of Cantor's theory of infinite sets to make the
following remarks a restatement of familiar material;  but if not, I
hope that they can stand as a first, and very partial, introduction to
these ideas.   The point about the first three examples is that we
can re-index the families of sets involved as simple sequences of sets.
For the first one,
this is elementary;  write $E'_n=E_{n+4}$ for $n\in\Bbb N$, and see that
$\bigcup_{n\ge 4}E_n=\bigcup_{n\in\Bbb N}E'_n\in\Sigma$.
For the other
two, we need to know something about the sets $\Bbb Z$ and
$\Bbb Q$.   We can find sequences $\langle k_n\rangle_{n\in\Bbb N}$ of
integers, and $\langle q_n\rangle_{n\in\Bbb N}$ of rational numbers,
such that every integer appears (at least once) as a $k_n$, and
every rational number appears (at least once) as a $q_n$;  that is, the
functions $n\mapsto k_n:\Bbb N\to\Bbb Z$ and
$n\mapsto q_n:\Bbb N\to\Bbb Q$ are surjective.   \prooflet{\Prf\ There are many ways of doing this;  one is to set

$$\eqalign{k_n&=\Bover{n}2\text{ for even }n,\cr
&=-\Bover{n+1}2\text{ for odd }n,\cr
q_n&=\Bover{n-m^3-m^2}{m+1}
  \text{ if }m\in\Bbb N\text{ and }m^3\le n<(m+1)^3.\cr}$$

\noindent (You should check carefully that these formulae do indeed do
what I claim they do.) \QeD}
Now, to deal with $\bigcup_{n\in\Bbb Z}E_n$, we can set

\Centerline{$E'_n=E_{k_n}\in\Sigma$}

\noindent for $n\in\Bbb N$, so that

\Centerline{$\bigcup_{n\in\Bbb Z}E_n=\bigcup_{n\in\Bbb N}E_{k_n}
=\bigcup_{n\in\Bbb N}E'_n\in\Sigma$,}

\noindent while for the other case we have

\Centerline{$\bigcup_{q\in\Bbb Q}E_q
=\bigcup_{n\in\Bbb N}E_{q_n}\in\Sigma$.}

Note that the first case $\bigcup_{n\ge 4}E_n$ can be thought of
as an
application of the same principle;  the map $n\mapsto n+4$ is a
surjection from $\Bbb N$ onto $\{4,5,6,7,\ldots\}$.
}%end of comment

\vleader{48pt}{111F}{Countable sets (a)}\cmmnt{ The common feature of the
sets
$\{n:n\ge 4\}$, $\Bbb Z$ and
$\Bbb Q$ which makes this procedure possible is that they are `countable'.
For our purposes here, the most natural definition of
countability is the following:} \dvro{A}{a} set $K$ is {\bf countable} if
either it is empty or there is a surjection from $\Bbb N$ onto $K$.   In
this case, if $\Sigma$ is a $\sigma$-algebra of sets and
$\langle  E_k\rangle_{k\in K}$ is a family in $\Sigma$ indexed by $K$, then
$\bigcup_{k\in K}E_k\in\Sigma$.   \prooflet{\Prf\ For if
$n\mapsto k_n:\Bbb N\to K$ is a surjection, then $E'_n=E_{k_n}\in\Sigma$ for every
$n\in\Bbb N$, and $\bigcup_{k\in K}E_k=\bigcup_{n\in\Bbb N}E'_n\in\Sigma$.   This
leaves out the case $K=\emptyset$;  but in this case the natural
interpretation of $\bigcup_{k\in K}E_k$ is

\Centerline{$\{x:\exists\enskip k\in \emptyset,\,x\in E_k\}$}

\noindent which is itself $\emptyset$, and therefore belongs to $\Sigma$
by clause (i) of 111A.\ \QeD}   \cmmnt{(In a sense this treatment of
$\emptyset$
is a conventional matter;
but there are various contexts in which we shall wish to discuss
$\bigcup_{k\in K}E_k$ without checking whether $K$ actually has any
members, and we need to be clear about what we will do in such cases.)}

\header{111Fb}{\bf (b)}\cmmnt{ There is an extensive, and enormously
important, theory concerning countable sets.   The only fragments which
I think we must have explicit at this point are the following.
(In \S1A1 I add a few words to link this presentation to conventional
approaches.)

\medskip

\quad}{\bf (i)}  If $K$ is countable and
$L\subseteq K$, then $L$ is countable.   \prooflet{\Prf\ If
$L=\emptyset$, this is
immediate.   Otherwise, take any $l^*\in L$, and a surjection
$n\mapsto k_n:\Bbb N\to K$ (of course $K$ also is not empty, as $l^*\in K$);  set
$l_n=k_n$ if $k_n\in L$, $l^*$ otherwise;  then $n\mapsto l_n:\Bbb N\to
L$ is a surjection.\ \Qed}

\medskip

\quad{\bf (ii)} The Cartesian product $\Bbb N\times\Bbb N=\{(m,n):m$,
$n\in\Bbb N\}$ is countable.   \prooflet{\Prf\ For each $n\in\Bbb N$,
let $k_n$,
$l_n\in\Bbb N$ be such that $n+1=2^{k_n}(2l_n+1)$;  that is, $k_n$ is
the power of $2$ in the prime factorisation of $n+1$, and $2l_n+1$ is
the (necessarily odd) number $(n+1)/2^{k_n}$.   Now
$n\mapsto(k_n,l_n)$ is a
surjection from $\Bbb N$ to $\Bbb N\times\Bbb N$.\ \Qed}\
\cmmnt{It will be
important to us later to know that $n\mapsto(k_n,l_n)$ is actually a
bijection, as is readily checked.}

\medskip

\quad{\bf (iii)} It follows that if $K$ and $L$ are countable sets, so
is $K\times L$.   \prooflet{\Prf\ If either $K$ or $L$ is empty, so is
$K\times
L$, so in this case $K\times L$ is certainly countable.   Otherwise, let
$\phi:\Bbb N\to K$ and $\psi:\Bbb N\to L$ be surjections;  then we have
a surjection $\theta:\Bbb N\times\Bbb N\to K\times L$ defined by setting
$\theta(m,n)=(\phi(m),\psi(n))$ for all $m$, $n\in\Bbb N$.   Now we know
from (ii) just above that there is also a surjection
$\chi:\Bbb N\to\Bbb N\times\Bbb N$, so that
$\theta\chi:\Bbb N\to K\times L$ is a
surjection, and $K\times L$ must be countable.\ \Qed}

\medskip

\quad{\bf (iv)}\dvro{ If}{ An induction on $r$ now shows
us that if} $K_1$, $K_2,\ldots,K_r$ are countable sets, so is
$K_1\times\ldots\times K_r$.   In particular,\cmmnt{ such sets as}
$\Bbb Q^r\times\Bbb Q^r$ will be countable, for any integer $r\ge 1$.


\header{111Fc}{\bf (c)}\dvro{ If}{ Putting 111Dd above together with
these
ideas, we see that if} $\Sigma$ is a $\sigma$-algebra of sets, $K$ is a
non-empty countable
set, and $\langle E_k\rangle_{k\in K}$ is a family in $\Sigma$, then

\Centerline{$\bigcap_{k\in K}E_k=\{x:x\in E_k\Forall k\in
K\}$}

\noindent belongs to $\Sigma$.   \prooflet{\Prf\ Let
$n\mapsto k_n:\Bbb N\to K$ be
a surjection;  then
$\bigcap_{k\in K}E_k=\bigcap_{n\in\Bbb N}E_{k_n}\in\Sigma$, as in 111Dd.\ \Qed}

\cmmnt{Note that there is a difficulty with the notion of
$\bigcap_{k\in
K}E_k$ if $K=\emptyset$;  the natural interpretation of this formula is
to read it as the universal class.   So ordinarily, when there is any
possibility that $K$ might be empty, one needs some such formulation as
$X\cap\bigcap_{k\in K}E_k$.
}%end of comment

\cmmnt{\header{111Fd}{\bf (d)} As an example of the way in which these
ideas will be used, consider the
following.   Suppose that $X$ is a set, $\Sigma$ is a $\sigma$-algebra
of subsets of $X$, and $\langle E_{qn}\rangle_{q\in\Bbb Q,n\in\Bbb N}$
is a family in $\Sigma$.   Then

\Centerline{$E
=\bigcap_{q\in\Bbb Q,q<\sqrt{2}}\bigcup_{m\in\Bbb N}\bigcap_{n\ge m}E_{qn}
=\bigcap_{q\in\Bbb Q,q<\sqrt{2}}(\bigcup_{m\in\Bbb N}(\bigcap_{n\ge m}E_{qn}))
\in\Sigma$.}

\noindent\vthsp\prooflet{\Prf\ Set
$F_{qm}=\bigcap_{n\ge m}E_{qn}=\bigcap_{n\in\Bbb N}E_{q,m+n}$ for $q\in\Bbb Q$
and $m\in\Bbb N$;
then every $F_{qm}$ belongs to $\Sigma$, by 111Dd or (c) above.   Set
$G_q=\bigcup_{m\in\Bbb N}F_{qm}$ for $q\in\Bbb Q$;  then every $G_q$
belongs to $\Sigma$, by 111A(iii).   Set
$K=\{q:q\in\Bbb Q,\,q<\sqrt{2}\}$;
then $K$ is countable, by 111E and (b-i) of this paragraph.   So
$\bigcap_{q\in K}G_q$ belongs to $\Sigma$, by (c).   But
$E=\bigcap_{q\in K}G_q$.\ \Qed}
}%end of comment

\cmmnt{\header{111Fe}{\bf (e)} And one final remark, which I give
without proof
here -- though many proofs will be implicit in the work below, and I
spell one out in 1A1Ha --

\medskip

\centerline{\bf The set $\Bbb R$ of real numbers is not countable.}

\medskip

\noindent So you must resist any temptation to look for a list $a_0$,
$a_1,\ldots$ running over the whole set of real numbers.
}%end of comment

\leader{111G}{Borel sets }\cmmnt{I can describe here one type of
non-trivial
$\sigma$-algebra;  the formulation is rather abstract, but the technique
is important
and the terminology is part of the basic vocabulary of measure theory.

\medskip

}{\bf (a)} Let $X$ be a set, and let $\frak S$ be any
non-empty family of
$\sigma$-algebras of subsets of $X$.   \cmmnt{(Thus a {\it member} of
$\frak S$ is itself a {\it family} of sets;
$\frak S\subseteq\Cal P(\Cal PX)$.)}   Then

\Centerline{$\bigcap\frak S=\{E:E\in\Sigma$ for every $\Sigma\in\frak
S\}$,}

\noindent the intersection of all the $\sigma$-algebras belonging to
$\frak S$, is a $\sigma$-algebra of subsets of $X$.   \prooflet{\Prf\
(i) By hypothesis, $\frak S$ is not empty;  take $\Sigma_0\in\frak S$;
then $\bigcap\frak S\subseteq\Sigma_0\subseteq\Cal PX$, so every member
of $\bigcap\frak S$ is a subset of $X$.   (ii) $\emptyset\in\Sigma$ for
every $\Sigma\in\frak S$,
so $\emptyset\in\bigcap\frak S$.   (iii) If
$E\in\bigcap\frak S$ then $E\in\Sigma$ for every $\Sigma\in\frak S$, so
$X\setminus E\in\Sigma$ for every $\Sigma\in\frak S$ and $X\setminus
E\in\bigcap\frak S$.   (iv) Let $\langle E_n\rangle_{n\in\Bbb N}$ be any
sequence in $\bigcap\frak S$.   Then for every $\Sigma\in\frak S$,
$\langle E_n\rangle_{n\in\Bbb N}$ is a sequence in $\Sigma$, so
$\bigcup_{n\in\Bbb N}E_n\in\Sigma$;  as $\Sigma$ is arbitrary,
$\bigcup_{n\in\Bbb N}E_n\in\bigcap\frak S$.\ \Qed}

\header{111Gb}{\bf (b)} Now let $\Cal A$ be any family of subsets of
$X$.   Consider

\Centerline{$\frak S=\{\Sigma:\Sigma$ is a $\sigma$-algebra of subsets
of $X$, $\Cal A\subseteq\Sigma\}$.}

\noindent\cmmnt{By definition, $\frak S$ is a family of
$\sigma$-algebras of
subsets of $X$;  also, it is not empty, because $\Cal PX\in\frak S$.
So} $\Sigma_{\Cal A}=\bigcap\frak S$ is a $\sigma$-algebra of subsets of
$X$\cmmnt{.   Because $\Cal A\subseteq\Sigma$ for every
$\Sigma\in\frak S$, $\Cal A\subseteq\Sigma_{\Cal A}$;  thus
$\Sigma_{\Cal A}$ itself
belongs to $\frak S$};  it is the smallest $\sigma$-algebra of subsets
of $X$ including $\Cal A$.

We say that $\Sigma_{\Cal A}$ is the $\sigma$-algebra of subsets of $X$
{\bf generated by} $\Cal A$.

\medskip

\noindent{\bf Examples (i)} For any $X$, the $\sigma$-algebra of subsets
of $X$ generated by $\emptyset$ is $\{\emptyset,X\}$.

\medskip

\quad{\bf (ii)} The
$\sigma$-algebra of subsets of $\Bbb N$ generated by
$\{\{n\}:n\in\Bbb N\}$ is $\Cal P\Bbb N$.

\header{111Gc}{\bf (c)(i)} We say that a set $G\subseteq\Bbb R$ is {\bf
open} if for
every $x\in G$ there is a $\delta>0$ such that the open interval
$\ooint{x-\delta,x+\delta}$ is included in $G$.

\medskip

\quad{\bf (ii)} Similarly, for any $r\ge 1$, we say that a set
$G\subseteq\BbbR^r$ is {\bf open} in $\BbbR^r$ if for every $x\in G$
there is a $\delta>0$ such that $\{y:\|y-x\|<\delta\}\subseteq G$,
where for $z=(\zeta_1,\ldots,\zeta_r)\in\BbbR^r$ I write
$\|z\|=\sqrt{\sum_{i=1}^r|\zeta_i|^2}$\cmmnt{;  thus $\|y-x\|$ is just
the ordinary Euclidean distance from $y$ to $x$}.

\header{111Gd}{\bf (d)} Now the {\bf Borel sets} of $\Bbb R$, or of
$\BbbR^r$, are
just the members of the $\sigma$-algebra of subsets of $\Bbb R$ or
$\BbbR^r$ generated by the family of open sets of $\Bbb R$ or $\BbbR^r$;
the $\sigma$-algebra itself is called the {\bf Borel $\sigma$-algebra}
in each case.

\cmmnt{\header{111Ge}{\bf (e)} Some readers will rightly feel that the
development here gives
very little idea of what a Borel set is `really' like.   (Open sets are
much easier;  see 111Ye.)   In fact the
importance of the concept derives largely from the fact that there are
alternative, more explicit, and in a sense more concrete, ways of
describing Borel sets.   I shall return to this topic in Chapter 42 in Volume 4.
}%end of comment

\exercises{
\leader{111X}{Basic exercises $\pmb{>}$(a)} Practise the algebra of
infinite unions and
intersections until you can confidently interpret such formulae as

\Centerline{$E\cap(\bigcup_{n\in\Bbb N}F_n)$,
  \qquad$\bigcup_{n\in\Bbb N}(E_n\setminus F)$,
  \qquad$E\cup(\bigcap_{n\in\Bbb N}F_n)$,}

\Centerline{$\bigcup_{n\in\Bbb N}(E\setminus F_n)$,
  \qquad$E\setminus(\bigcup_{n\in\Bbb N}F_n)$,
  \qquad$\bigcap_{n\in\Bbb N}(E_n\setminus F)$,}

\Centerline{$E\setminus(\bigcap_{n\in\Bbb N}F_n)$,
  \qquad$\bigcap_{n\in\Bbb N}(E\cup F_n)$,
  \qquad$(\bigcup_{n\in\Bbb N}E_n)\setminus F$,}

\Centerline{$\bigcup_{n\in\Bbb N}(E\cap F_n)$,
  \qquad$(\bigcap_{n\in\Bbb N}E_n)\setminus F$,
  \qquad$\bigcap_{n\in\Bbb N}(E\setminus F_n)$,}

\Centerline{$(\bigcup_{n\in\Bbb N}E_n)\cap(\bigcup_{n\in\Bbb N}F_n)$,
  \qquad$\bigcap_{m,n\in\Bbb N}(E_m\setminus F_n)$,
  \qquad$(\bigcap_{n\in\Bbb N}E_n)\cup(\bigcap_{n\in\Bbb N}F_n)$,}

\Centerline{$\bigcap_{m,n\in\Bbb N}(E_m\cup F_n)$,
  \qquad$(\bigcap_{n\in\Bbb N}E_n)\setminus(\bigcup_{n\in\Bbb N}F_n)$,
  \qquad$\bigcup_{m,n\in\Bbb N}(E_m\cap F_n)$,}

\noindent and, in particular, can identify the nine pairs into which
these formulae naturally fall.

\sqheader 111Xb In $\Bbb R$, show that all `open intervals'
$\ooint{a,b}$, $\ooint{-\infty,b}$, $\ooint{a,\infty}$ are open sets,
and that all intervals (bounded or unbounded, open, closed or half-open)
are Borel sets.

\sqheader 111Xc Let $X$ and $Y$ be sets and $\Sigma$ a
$\sigma$-algebra of subsets of $X$.   Let $\phi:X\to Y$ be a function.
Show that $\{F:F\subseteq Y,\,\phi^{-1}[F]\in\Sigma\}$ is a
$\sigma$-algebra of subsets of $Y$.   (See 1A1B for the notation here.)

\sqheader 111Xd Let $X$ and $Y$ be sets and $\Tau$ a
$\sigma$-algebra of subsets of $Y$.   Let $\phi:X\to Y$ be a function.
Show that $\{\phi^{-1}[F]:F\in\Tau\}$ is a
$\sigma$-algebra of subsets of $X$.

\spheader 111Xe Let $X$ be a set, $\Cal A$ a family of subsets of $X$,
and $\Sigma$ the $\sigma$-algebra of subsets of $X$ generated by $\Cal
A$.   Suppose that $Y$ is another set and $\phi:Y\to X$ a function.   Show
that $\{\phi^{-1}[E]:E\in\Sigma\}$ is the $\sigma$-algebra of subsets of
$Y$ generated by $\{\phi^{-1}[A]:A\in\Cal A\}$.

\spheader 111Xf Let $X$ be a set, $\Cal A$ a family of subsets of $X$,
and $\Sigma$ the $\sigma$-algebra of subsets of $X$ generated by $\Cal
A$.   Suppose that $Y\subseteq X$.   Show that $\{E\cap Y:E\in\Sigma\}$
is the $\sigma$-algebra of subsets of $Y$ generated by 
$\{A\cap Y:A\in\Cal A\}$.

\leader{111Y}{Further exercises (a)} In $\BbbR^r$, where $r\ge 1$, show
that $G+a=\{x+a:x\in G\}$ is open whenever $G\subseteq\BbbR^r$ is open
and $a\in\BbbR^r$.   Hence show that $E+a$ is a Borel set whenever
$E\subseteq\BbbR^r$ is a Borel set and $a\in\BbbR^r$.    \Hint{show that
$\{E:E+a\text{ is a Borel set}\}$ is a $\sigma$-algebra
containing all open sets.}

\header{111Yb}{\bf (b)} Let $X$ be a set, $\Sigma$ a $\sigma$-algebra of
subsets of $X$ and $A$ any subset of $X$.   Show that $\{(E\cap
A)\cup(F\setminus A):E,\,F\in\Sigma\}$ is a $\sigma$-algebra of subsets
of $X$, the $\sigma$-algebra generated by $\Sigma\cup\{A\}$.

\header{111Yc}{\bf (c)} Let $G\subseteq\BbbR^2$ be an open set.   Show
that all the horizontal and vertical sections

\Centerline{$\{\xi:(\xi,\eta)\in G\}$,
\quad$\{\xi:(\eta,\xi)\in G\}$}

\noindent of $G$ are open subsets of $\Bbb R$.

\header{111Yd}{\bf (d)} Let $E\subseteq\BbbR^2$ be a Borel set.   Show
that all the horizontal and vertical sections

\Centerline{$\{\xi:(\xi,\eta)\in E\}$,
\quad$\{\xi:(\eta,\xi)\in E\}$}

\noindent of $E$ are Borel subsets of $\Bbb R$.   \Hint{show
that the family of subsets of $\BbbR^2$ whose sections are all Borel
sets is a $\sigma$-algebra of subsets of $\BbbR^2$ containing all the
open sets.}

\spheader 111Ye Let $G\subseteq\Bbb R$ be an open set.   Show that $G$
is uniquely expressible as the union of a countable (possibly empty)
family $\Cal I$ of open intervals (the `components' of $G$) no two of
which have any point in common.   \Hint{for $x$,
$y\in G$ say that $x\sim y$ if every point between $x$ and $y$ belongs
to $G$.   Show that $\sim$ is an equivalence relation.   Let $\Cal I$ be
the set of equivalence classes.}
}%end of exercises

\endnotes{
\Notesheader{111} I suppose that the most important concept in this
section is the one introduced tangentially in 111E-111F, the idea of
`countable' set.   While it is possible to avoid much of the formal
theory of infinite sets for the time being, I do not think it is
possible to make sense of this chapter without a firm notion of the
difference between `finite' and `infinite', and some intuitions
concerning `countability'.   In particular, you must remember that
infinite sets are not, in general, countable, and that $\sigma$-algebras
are not, in general, closed under arbitrary unions.

The next thing to be sure of is that you can cope with the set-theoretic
manipulations here, so that such formulae as
$\bigcap_{n\in\Bbb N}E_n=X\setminus\bigcup_{n\in\Bbb N}(X\setminus E_n)$ (111Dd) are, if
not yet transparent, at least not alarming.   A large proportion of
the volume will be expressed in this language, and reasonable fluency is
essential.

Finally, for those who are looking for an actual idea to work on
straight away, I offer the concept of $\sigma$-algebra `generated' by a
collection $\Cal A$ (111G).   The point of the definition here is that
it involves consideration of a family $\frak S\in\Cal P(\Cal P(\Cal PX))$,
even though both $\Cal A$ and $\Sigma_{\Cal A}$ are subsets of
$\Cal PX$;  we need to work a layer or two up in the hierarchy of power sets.
You may have seen, for instance, the concept of `linear subspace $U$
generated by vectors $u_1,\ldots,u_n$'.   This can be defined as the
intersection of all the linear subspaces containing the vectors
$u_1,\ldots,u_n$, which is the method corresponding to that of 111Ga-b;
but it also has an `internal' definition, as the set of vectors
expressible as $\alpha_1u_1+\ldots+\alpha_nu_n$ for scalars $\alpha_i$.
For $\sigma$-algebras, however, there is no such simple `internal'
definition available (though there is a great deal to be said in this
direction which I think we are not yet ready for;  some ideas may be
found in \S136).   This is primarily
because of (iii) in the definition 111A;  a $\sigma$-algebra must be
closed under an infinitary operation, that is, the operation of union
applied to infinite sequences of sets.   By contrast, a linear subspace
of a vector space need be closed only under the finitary operations of
scalar multiplication and addition, each involving at most two vectors
at a time.
}%end of notes

\discrpage


\frfilename{mt112.tex}
\versiondate{20.2.05/20.8.08}

\def\chaptername{Measure spaces}
\def\sectionname{Measure spaces}
\copyrightdate{1994}

\newsection{112}

We are now, I hope, ready for the second major definition, the
definition on which all the work of this treatise is based.

\leader{112A}{Definition} A {\bf measure space} is a triple
$(X,\Sigma,\mu)$ where

\quad(i) $X$ is a set;

\quad(ii) $\Sigma$ is a $\sigma$-algebra of subsets of $X$;

\quad(iii) $\mu:\Sigma\to[0,\infty]$ is a function such that

\qquad($\alpha$) $\mu\emptyset=0$;

\qquad($\beta$) if $\langle E_n\rangle_{n\in\Bbb N}$ is a disjoint
sequence in $\Sigma$, then
$\mu(\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\mu E_n$.

\noindent In this context, members of $\Sigma$ are called {\bf
measurable} sets, and $\mu$ is called a {\bf measure on $X$}.

\leader{112B}{Remarks}\cmmnt{ {\bf (a) The use of $\infty$} In (iii) of
the definition
above, I declare that $\mu$ is to be a function taking values in
`$[0,\infty]$', that is, the set comprising the non-negative real
numbers with `$\infty$' adjoined.   I expect that you have already
encountered various uses of the symbol $\infty$ in analysis;   I hope
you have realised that it means rather different things in different
contexts, and that it is necessary to establish clear conventions for
its use each time.   The `$\infty$ of measure' corresponds to the
notion of infinite length or area or volume.   The basic operation we
need to perform on it is addition:  $\infty+a=a+\infty=\infty$ for every
$a\in[0,\infty\roibr$ (that is, every real number $a\ge 0$), and
$\infty+\infty=\infty$.   This renders $[0,\infty]$ a semigroup under
addition.   It will be reasonably safe to declare
$\infty-a=\infty$ for every $a\in\Bbb R$;  but we must
absolutely decline to interpret the formula $\infty-\infty$.   As for
multiplication, it turns out that it is usually right to interpret
$\infty\cdot\infty$, $a\cdot\infty$ and $\infty\cdot a$ as $\infty$ for
$a>0$, while $0\cdot\infty=\infty\cdot 0$ can generally be taken as $0$.

We also have a natural total ordering of $[0,\infty]$, writing
$a<\infty$ for every $a\in\coint{0,\infty}$.   This gives an idea of
supremum and infimum of an arbitrary (non-empty) subset of $[0,\infty]$;
and it will often be right to interpret $\inf\emptyset$ as $\infty$, but
I will try to signal this particular convention each time it is
relevant.
We also have a notion of limit;  if
$\langle u_n\rangle_{n\in\Bbb N}$ is a sequence in $[0,\infty]$, then it converges to $u\in[0,\infty]$
if

\qquad {for every $v<u$ there is an $n_0\in\Bbb N$ such that $v\le
u_n$ for every $n\ge n_0$,}

\qquad {for every $v>u$ there is an $n_0\in\Bbb N$ such that $v\ge
u_n$ for every $n\ge n_0$.}

\noindent Of course if $u=0$ or $u=\infty$ then one of these clauses
will be vacuously satisfied.

(See also \S135.)

\header{112Bb}{\bf (b)} I should say plainly what I mean by a `disjoint'
sequence:
a sequence $\langle E_n\rangle_{n\in\Bbb N}$ is {\bf disjoint} if no
point belongs to more than one $E_n$, that is, if $E_m\cap
E_n=\emptyset$ for all distinct $m$, $n\in\Bbb N$.   Note that there is
no bar here on one, or many, of the $E_n$ being the empty set.

Similarly, if $\langle E_i\rangle_{i\in I}$ is a family of sets indexed
by an arbitrary set $I$, it is {\bf disjoint} if $E_i\cap E_j=\emptyset$
for all distinct $i$, $j\in I$.

\medskip

} {\bf (c)} In interpreting clause (iii-$\beta$) of the
definition above,
we need to assign values to sums $\sum_{n=0}^{\infty}u_n$ for arbitrary
sequences $\langle u_n\rangle_{n\in\Bbb N}$ in $[0,\infty]$.
\cmmnt{The
natural way to do this is to say that
$\sum_{n=0}^{\infty}u_n=\lim_{n\to\infty}\sum_{m=0}^nu_m$,
using the definitions sketched in (a).} %end of comment
If one of the $u_m$ is itself infinite, \cmmnt{say $u_k=\infty$,
then $\sum_{m=0}^nu_m=\infty$ for every
$n\ge k$, so of course} $\sum_{n=0}^{\infty}u_n=\infty$.
If all the $u_m$ are finite, then\cmmnt{, because they are all
non-negative,} the sequence $\langle\sum_{m=0}^nu_m\rangle_{n\in\Bbb N}$
of partial
sums is monotonic non-decreasing, and either has a finite limit
$\sum_{n=0}^{\infty}u_n\in\Bbb R$, or diverges to $\infty$;  in which
case we again interpret $\sum_{n=0}^{\infty}u_n$ as $\infty$.

\header{112Bd}{\bf (d)}\cmmnt{ Once again, the important examples of
measure spaces will have
to wait until \S\S114 and 115 below.   However, I can describe
immediately one particular class of measure space, which should always
be borne in
mind, though it does not give a good picture of the most important and
interesting parts of the subject.}   Let $X$ be any set, and let
$h:X\to[0,\infty]$ be any function.   For every $E\subseteq X$ write
$\mu E=\sum_{x\in E}h(x)$.   To interpret this sum, note that there is
no difficulty for
finite sets $E$ (taking $\sum_{x\in\emptyset}h(x)=0$), while for
infinite sets $E$ we can take
$\sum_{x\in E}h(x)=\sup\{\sum_{x\in I}h(x):I\subseteq E$ is finite$\}$,
because
every $h(x)$ is non-negative.   \cmmnt{(You may well prefer to think
about this at first with $X=\Bbb N$, so that
$\sum_{n\in E}h(n)=\lim_{n\to\infty}\sum_{m\in E,m\le n}h(m)$;
but I hope that a little thought will show you that the
general case, in which $X$ may even be uncountable, is not really more
difficult.)}   Now $(X,\Cal PX,\mu)$ is a measure space.

\cmmnt{We are very far from being ready for the specialized vocabulary
needed to
describe different kinds of measure space, but when the time comes} I will call
measures of this kind {\bf point-supported}.

Two particular cases recur often enough to be worth giving names to.
If $h(x)=1$ for every $x$, then $\mu E$
is just the number of points of $E$ if $E$ is finite, and is $\infty$ if
$E$ is infinite.   I will call this {\bf counting measure} on $X$.
If $x_0\in X$, we can set $h(x_0)=1$ and $h(x)=0$ for
$x\in X\setminus\{x_0\}$;  then $\mu E$ is $1$ if $x_0\in E$, and $0$ for
other $E$.   I will call this the
{\bf Dirac measure on $X$ concentrated at $x_0$}.
Another simple example is with $X=\Bbb N$, $h(n)=2^{-n-1}$ for
every $n$;  then $\mu X={1\over 2}+{1\over 4}+\ldots=1$.

\spheader 112Be If $(X,\Sigma,\mu)$ is a measure space\cmmnt{, then
$\Sigma$ is the domain of the function $\mu$, and $X$ is the largest
member of $\Sigma$.   We can therefore recover the whole triplet
$(X,\Sigma,\mu)$ from its final component $\mu$.   This is not a game
which is worth playing at this stage.   However, it is convenient on
occasion to introduce a measure without immediately giving a name to its
domain, and when I do this} I may say that `$\mu$ {\bf measures} $E$' or
`$E$ is {\bf measured by} $\mu$' to mean that $\mu E$ is
defined\cmmnt{, that is, that $E$ belongs to the $\sigma$-algebra
$\dom\mu$}.   \cmmnt{{\bf Warning!} Many authors use the phrase
`$\mu$-measurable set' to mean something a little different from what I am
discussing here.}

\leader{112C}{Elementary properties of measure spaces} Let
$(X,\Sigma,\mu)$ be a measure space.

(a) If $E$, $F\in\Sigma$ and $E\cap F=\emptyset$ then $\mu(E\cup F)=\mu
E+\mu F$.

(b) If $E$, $F\in\Sigma$ and $E\subseteq F$ then $\mu E\le \mu F$.

(c) $\mu(E\cup F)\le\mu E+\mu F$ for any $E$, $F\in\Sigma$.

(d) If $\langle E_n\rangle_{n\in\Bbb N}$ is any sequence in $\Sigma$,
then $\mu(\bigcup_{n\in\Bbb N}E_n)\le\sum_{n=0}^{\infty}\mu E_n$.

(e) If $\langle E_n\rangle_{n\in\Bbb N}$ is a non-decreasing sequence in
$\Sigma$ (that is, $E_n\subseteq E_{n+1}$ for every $n\in\Bbb N$) then

\Centerline{$\mu(\bigcup_{n\in\Bbb N}E_n)=\lim_{n\to\infty}\mu
E_n=\sup_{n\in\Bbb N}\mu E_n$.}

(f) If $\langle E_n\rangle_{n\in\Bbb N}$ is a non-increasing sequence in
$\Sigma$ (that is, $E_{n+1}\subseteq E_{n}$ for every $n\in\Bbb N$), and
if some $\mu E_n$ is finite, then

\Centerline{$\mu(\bigcap_{n\in\Bbb N}E_n)
=\lim_{n\to\infty}\mu E_n=\inf_{n\in\Bbb N}\mu E_n$.}

\proof{{\bf (a)} Set $E_0=E$, $E_1=F$, $E_n=\emptyset$ for $n\ge
2$;  then $\langle E_n\rangle_{n\in\Bbb N}$ is a disjoint sequence in
$\Sigma$ and $\bigcup_{n\in\Bbb N}E_n=E\cup F$, so

\Centerline{$\mu(E\cup
F)=\sum_{n=0}^{\infty}\mu E_n=\mu E+\mu F$}

\noindent (because $\mu\emptyset=0$).

\medskip

{\bf (b)} $F\setminus E\in\Sigma$ (111Dc) and
$\mu(F\setminus E)\ge 0$ (because all values of $\mu$ are in $[0,\infty]$);
so (using (a))

\Centerline{$\mu F=\mu E+\mu(F\setminus E)\ge\mu E$.}

\medskip

{\bf (c)} $\mu(E\cup F)=\mu E+\mu(F\setminus E)$, by (a), and
$\mu(F\setminus E)\le \mu F$, by (b).

\medskip

{\bf (d)} Set $F_0=E_0$, $F_n=E_n\setminus\bigcup_{i<n}E_i$ for $n\ge
1$;  then $\langle F_n\rangle_{n\in\Bbb N}$ is a disjoint sequence in
$\Sigma$, $\bigcup_{n\in\Bbb N}F_n=\bigcup_{n\in\Bbb N}E_n$ and
$F_n\subseteq E_n$ for every $n$.   By (b) just above, $\mu F_n\le\mu
E_n$ for each $n$;  so

\Centerline{$\mu(\bigcup_{n\in\Bbb N}E_n)
=\mu(\bigcup_{n\in\Bbb N}F_n)=\sum_{n=0}^{\infty}\mu
F_n\le\sum_{n=0}^{\infty}\mu E_n$.}

\medskip

{\bf (e)} Set $F_0=E_0$, $F_n=E_n\setminus E_{n-1}$ for $n\ge 1$;  then
$\langle F_n\rangle_{n\in\Bbb N}$ is a disjoint sequence in $\Sigma$ and
$\bigcup_{n\in\Bbb N}F_n=\bigcup_{n\in\Bbb N}E_n$.   Consequently
$\mu(\bigcup_{n\in\Bbb N}E_n)=\sum_{n=0}^{\infty}\mu F_n$.   But an easy
induction on $n$, using (a) for the inductive step, shows that $\mu
E_n=\sum_{m=0}^n\mu F_m$ for every $n$.   So

\Centerline{$\sum_{n=0}^{\infty}\mu
F_n=\lim_{n\to\infty}\sum_{m=0}^n\mu F_m=\lim_{n\to\infty}\mu E_n$.}

\noindent Finally, $\lim_{n\to\infty}\mu E_n=\sup_{n\in\Bbb N}\mu E_n$
because (by
(b)) $\langle\mu E_n\rangle_{n\in\Bbb N}$ is non-decreasing.

\medskip

{\bf (f)} Suppose that $\mu E_k<\infty$.   Set $F_n=E_k\setminus
E_{k+n}$ for $n\in\Bbb N$, $F=\bigcup_{n\in\Bbb N}F_n$;  then $\langle
F_n\rangle_{n\in\Bbb N}$ is a non-decreasing sequence in $\Sigma$, so
$\mu
F=\lim_{n\to\infty}\mu F_n$, by (e) just above.   Also, $\mu F_n+\mu
E_{k+n}=\mu E_k$;  because $\mu E_k<\infty$, we may safely write $\mu
F_n=\mu E_k-\mu E_{k+n}$, so that

\Centerline{$\mu F=\lim_{n\to\infty}(\mu E_k-\mu
E_{k+n})=\mu E_k-\lim_{n\to\infty}\mu E_n$.}

\noindent   Next, $F\subseteq E_k$, so
$\mu F+\mu(E_k\setminus F)=\mu E_k$, and (again because $\mu E_k$ is
finite) $\mu F=\mu E_k-\mu(E_k\setminus F)$.   Thus we must have
$\mu(E_k\setminus F)=\lim_{n\to\infty}\mu E_n$.   But $E_k\setminus F$
is just $\bigcap_{n\in\Bbb N}E_n$.

Finally, $\lim_{n\to\infty}\mu E_n=\inf_{n\in\Bbb N}\mu E_n$ because
$\langle\mu E_n\rangle_{n\in\Bbb N}$ is non-increasing.
}%end of proof of 112C

\medskip

\cmmnt{\noindent{\bf Remark} Observe that in (f) above it is essential
to have
$\inf_{n\in\Bbb N}\mu E_n<\infty$.   The construction in 112Bd is
already enough to show this.   Take $X=\Bbb N$ and let $\mu$ be counting
measure on $X$.   Set $E_n=\{i:i\in\Bbb N,\,i\ge n\}$ for each $n$.
Then $E_{n+1}\subseteq E_n$ for each $n$, but

\Centerline{$\mu(\bigcap_{n\in\Bbb N}E_n)
=\mu\emptyset=0<\infty=\lim_{n\to\infty}\mu E_n$.}
}%end of comment


\leader{112D}{Negligible sets} Let $(X,\Sigma,\mu)$ be any measure
space.

\header{112Da}{\bf (a)} A set $A\subseteq X$ is {\bf negligible} (or
{\bf null}) if
there is a set $E\in\Sigma$ such that $A\subseteq E$ and
$\mu E=0$.   (If there seems to be a possibility of doubt about which 
measure is involved, I will write {\bf $\mu$-negligible}.)

\header{112Db}{\bf (b)} Let $\Cal N$ be the family of negligible subsets
of $X$.
Then (i) $\emptyset\in\Cal N$ (ii) if $A\subseteq B\in\Cal N$ then
$A\in\Cal N$ (iii) if $\langle A_n\rangle_{n\in\Bbb N}$ is any sequence
in $\Cal N$, $\bigcup_{n\in\Bbb N}A_n\in\Cal N$.   \prooflet{\Prf\ (i)
$\mu(\emptyset)=0$.   (ii) There is an $E\in\Sigma$ such that $\mu E=0$
and $B\subseteq E$;  now $A\subseteq E$.   (iii) For each $n\in\Bbb N$
choose an $E_n\in\Sigma$ such that $A_n\subseteq E_n$ and $\mu E_n=0$.
Now $E=\bigcup_{n\in\Bbb N}E_n\in\Sigma$ and
$\bigcup_{n\in\Bbb N}A_n\subseteq\bigcup_{n\in\Bbb N}E_n$, and
$\mu(\bigcup_{n\in\Bbb N}E_n)\le\sum_{n=0}^{\infty}\mu E_n$, by 112Cd,
so $\mu(\bigcup_{n\in\Bbb N}E_n)=0$.  \Qed}

I will call $\Cal N$ the {\bf null ideal} of the measure $\mu$.
(A family of sets satisfying the conditions (i)-(iii) here is called a
{\bf $\sigma$-ideal} of sets.)

\header{112Dc}{\bf (c)} A set $A\subseteq X$ is {\bf conegligible} if
$X\setminus A$
is negligible\cmmnt{;  that is, there is a measurable set
$E\subseteq A$
such that $\mu(X\setminus E)=0$}.   Note that (i) $X$ is conegligible
(ii) if $A\subseteq B\subseteq X$ and $A$ is conegligible then $B$ is
conegligible (iii) if $\langle A_n\rangle_{n\in\Bbb N}$ is a sequence of
conegligible sets, then $\bigcap_{n\in\Bbb N}A_n$ is conegligible.

\header{112Dd}{\bf (d)}\cmmnt{ It is convenient, and customary, to use
some relatively informal language concerning negligible sets.}   If
$P(x)$ is
some assertion applicable to members $x$ of the set $X$, we say that

\Centerline{`$P(x)$ for almost every $x\in X$'}

\noindent or

\Centerline{`$P(x)$ a.e.\ $(x)$'}

\noindent or

\Centerline{`$P$ almost everywhere',\quad `$P$ a.e.'}

\noindent or\cmmnt{, if it seems necessary to specify the measure
involved,}

\Centerline{`$P(x)$ for $\mu$-almost every $x$',
\quad`$P(x)\,\mu$-a.e.$(x)$',
\quad`$P\,\mu$-a.e.',}

\noindent to mean that

\Centerline{$\{x:x\in X,\,P(x)\}$}

\noindent is conegligible in $X$\cmmnt{, that is, that

\Centerline{$\{x:x\in X,\,P(x)$ is false$\}$}

\noindent is negligible}.   Thus\cmmnt{, for instance,} if
$f:X\to\Bbb R$ is a function, `$f>0$ a.e.' means that $\{x:f(x)\le 0\}$ is negligible.

\cmmnt{\spheader 112De The phrases `{\bf almost surely}' ({\bf
a.s.}), `{\bf presque partout}' ({\bf p.p.}) are also used for `almost
everywhere'.
}%end of comment

\cmmnt{\header{112Df}{\bf (f)} I should call your attention to the
fact that, on my
definitions, a negligible set need not itself be measurable, though it
must be included in some negligible measurable set.   (Measure spaces in
which all negligible sets are measurable are called
{\bf complete}.   I will return to this question in \S211.)
}%end of comment

\spheader 112Dg When $f$ and $g$ are real-valued functions defined on
conegligible subsets of a measure space, I will write $f\eae g$,
$f\leae g$ or $f\geae g$ to mean, respectively,

\Centerline{$f=g$ a.e., that is,
$\{x:x\in\dom(f)\cap\dom(g)$, $f(x)=g(x)\}$ is conegligible,}

\Centerline{$f\le g$ a.e., that is,
$\{x:x\in\dom(f)\cap\dom(g)$, $f(x)\le g(x)\}$ is conegligible,}

\Centerline{$f\ge g$ a.e., that is,
$\{x:x\in\dom(f)\cap\dom(g)$, $f(x)\ge g(x)\}$ is conegligible.}

\exercises{
\leader{112X}{Basic exercises $\pmb{>}$(a)}
%\sqheader 112Xa
Let $(X,\Sigma,\mu)$ be a
measure space.   Show that (i)
$\mu(E\cup F)+\mu(E\cap F)=\mu E+\mu F$ (ii)
$\mu(E\cup F\cup G)+\mu(E\cap F)+\mu(E\cap G)+\mu(F\cap G)=\mu E+\mu
F+\mu G+\mu(E\cap F\cap G)$ for all $E$, $F$, $G\in\Sigma$.   Generalize
these results to longer sequences of sets.   (You may prefer to begin
with the case in which $\mu E$, $\mu F$ and $\mu G$ are all finite.
But I hope you will be able to find arguments which deal with the
general case.)
%112C

\sqheader 112Xb Let $(X,\Sigma,\mu)$ be a measure space and
$\langle E_n\rangle_{n\in\Bbb N}$ a sequence in $\Sigma$.   Show that

\Centerline{$\mu(\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}E_m)
\le\liminf_{n\to\infty}\mu E_n$.}
%112C

\spheader 112Xc Let $(X,\Sigma,\mu)$ be a measure space, and
$E$, $F\in\Sigma$;  suppose that $\mu E<\infty$.   Show that
$\mu(F\symmdiff E)=\mu F-\mu E+2\mu(E\setminus F)$.
%112C

\spheader 112Xd Let $(X,\Sigma,\mu)$ be a measure space and
$\sequencen{E_n}$ a sequence of measurable sets such that
$\mu(\bigcup_{n\in\Bbb N}E_n)<\infty$.   (i) Show that
$\limsup_{n\to\infty}\mu E_n
\le\mu(\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}E_m)$.   (ii) Show that
if $\bigcap_{n\in\Bbb N}\bigcup_{m\ge n}E_m=E
=\bigcup_{n\in\Bbb N}\bigcap_{m\ge n}E_m$ then
$\lim_{n\to\infty}\mu E_n$ exists and is equal to $\mu E$.
%112Xb 112C

\sqheader 112Xe Let $(X,\Sigma,\mu)$ be a measure space, and $\eusm F$
the set of
real-valued functions whose domains are conegligible subsets of $X$.
(i) Show that $\{(f,g):f,\,g\in\eusm F,\,f\leae g\}$ and
$\{(f,g):f,\,g\in\eusm F,\,f\geae g\}$ are reflexive transitive relations
on $\eusm F$, each the inverse of the other.   (ii) Show that
$\{(f,g):f,\,g\in\eusm F,\,f\eae g\}$ is their intersection, and is an
equivalence relation on $\eusm F$.
%112D

\spheader 112Xf Let $(X,\Sigma,\mu)$ be a measure space, $Y$ a set, and
$\phi:X\to Y$ a function.   Set
$\Tau=\{F:F\subseteq Y$, $\phi^{-1}[F]\in\Sigma\}$ and
$\nu F=\mu\phi^{-1}[F]$ for $F\in\Tau$.   Show that $\nu$ is a measure on
$Y$.   ($\nu$ is called the {\bf image measure} on $Y$, and I will
generally denote it $\mu\phi^{-1}$.)
%111Xc

\leader{112Y}{Further exercises (a)}
%\spheader 112Ya
Let $X$ be a set and $\Sigma$ a
$\sigma$-algebra of subsets of $X$.   Let $\mu_1$ and $\mu_2$ be two
measures on $X$, both with domain $\Sigma$.   Set

\Centerline{$\mu E
=\inf\{\mu_1(E\cap F)+\mu_2(E\setminus F):F\in\Sigma\}$}

\noindent for each $E\in\Sigma$.   Show that $\mu$ is a measure on $X$,
and that it is the greatest measure, with domain $\Sigma$, such that
$\mu E\le\min(\mu_1E,\mu_2E)$ for every $E\in\Sigma$.
%problem with double use of symbol  \mu , here and in next exercise,
%see correspondence with Wallace Thompson

\spheader 112Yb Let $X$ be a set and $\Sigma$ a
$\sigma$-algebra of subsets of $X$.   Let $\mu_1$ and $\mu_2$ be two
measures on $X$, both with domain $\Sigma$.   Set

\Centerline{$\mu E
=\sup\{\mu_1(E\cap F)+\mu_2(E\setminus F):F\in\Sigma\}$}

\noindent for each $E\in\Sigma$.   Show that $\mu$ is a measure on $X$,
and that it is the least measure, with domain $\Sigma$, such that $\mu
E\ge\max(\mu_1E,\mu_2E)$ for every $E\in\Sigma$.

\spheader 112Yc Let $X$ be a set and $\Sigma$ a
$\sigma$-algebra of subsets of $X$.

\quad(i) Suppose that $\nu_0,\ldots,\nu_n$ are measures on $X$, all with domain
$\Sigma$.   Set

\Centerline{$\mu E=\inf\{\sum_{i=0}^n\nu_iF_i:F_0,\ldots,F_n\in\Sigma$,
$E\subseteq\bigcup_{i\le n}F_i\}$}

\noindent for $E\in\Sigma$.   Show that $\mu$ is a measure on $X$.

\quad(ii) Let $\Nu$ be a non-empty family of
measures on $X$, all with domain $\Sigma$.   Set

$$\eqalign{\mu E=\inf\{\sum_{n=0}^{\infty}\nu_nF_n:
&\sequencen{\nu_n}\text{ is a sequence in }\Nu,\cr
&\sequencen{F_n}\text{ is a sequence in }\Sigma,
\,E\subseteq\bigcup_{n\in\Bbb N}F_n\}\cr}$$

\noindent for $E\in\Sigma$.   Show that $\mu$ is a measure on $X$.

\quad(iii) Let $\Nu$ be a non-empty family of
measures on $X$, all with domain $\Sigma$, and suppose that there is some
$\nuprime\in\Nu$ such that $\nuprime X<\infty$.   Set

\Centerline{$\mu E=\inf\{\sum_{i=0}^n\nu_iF_i:n\in\Bbb N,
  \,\nu_0,\ldots,\nu_n\in\Nu,\,F_0,\ldots,F_n\in\Sigma,
  E\subseteq\bigcup_{i\le n}F_i\}$}

\noindent for $E\in\Sigma$.   Show that $\mu$ is a measure on $X$.

\quad(iv) Suppose, in (iii), that $\Nu$ is downwards-directed, that is,
for any $\nu_1$, $\nu_2\in\Nu$ there is a $\nu\in\Nu$ such that
$\nu E\le\min(\nu_1E,\nu_2E)$ for every $E\in\Sigma$.   Show that
$\mu E=\inf_{\nu\in\Nu}\nu E$ for every $E\in\Sigma$.

\quad(v) Show that in all the cases (i)-(iii) the measure constructed
is the greatest measure $\mu$ with domain $\Sigma$ such that
$\mu E\le\inf_{\nu\in\Nu}\nu E$ for every $E\in\Sigma$.

\spheader 112Yd Let $X$ be a set and $\Sigma$ a
$\sigma$-algebra of subsets of $X$.   Let $\Nu$ be a non-empty family of
measures on $X$, all with domain $\Sigma$.   Set

$$\eqalign{\mu E=\sup\{\sum_{i=0}^n\nu_iF_i:n\in\Bbb N,
  &\,\nu_0,\ldots,\nu_n\in\Nu,\cr
  &F_0,\ldots,F_n\text{ are disjoint subsets of }E
    \text{ belonging to }\Sigma\}\cr}$$

\noindent for $E\in\Sigma$.   (i) Show that

$$\eqalign{\mu E=\sup\{\sum_{n=0}^{\infty}\nu_nF_n:
&\sequencen{\nu_n}\text{ is a sequence in }\Nu,\cr
&\sequencen{F_n}\text{ is a disjoint sequence in }\Sigma,
\,\bigcup_{n\in\Bbb N}F_n\subseteq E\}\cr}$$

\noindent for every $E\in\Sigma$.
(ii) Show that $\mu$ is a measure on $X$, and that it is the
least measure, with domain $\Sigma$, such that
$\mu E\ge\sup_{\nu\in\Nu}\nu E$ for every $E\in\Sigma$.   (iii) Now
suppose that $\Nu$ is upwards-directed, that is, for any $\nu_1$,
$\nu_2\in\Nu$ there is a $\nu\in\Nu$ such that
$\nu E\ge\max(\nu_1E,\nu_2E)$ for every $E\in\Sigma$.   Show that
$\mu E=\sup_{\nu\in\Nu}\nu E$ for every $E\in\Sigma$.

\spheader 112Ye Let $(X,\Sigma,\mu)$ be a measure space and
$\sequencen{E_n}$ a sequence of measurable sets.   For each $k\in\Bbb N$
set $H_k=\{x:x\in X,\,\#(\{n:x\in E_n\})\ge k\}$, the set of points
belonging to $E_n$ for $k$ or more values of $n$.   (i) Show that each
$H_k$ is measurable.   (ii) Show that $\sum_{k=1}^{\infty}\mu
H_k=\sum_{n=0}^{\infty}\mu E_n$.   \Hint{start with the case in which
$E_n=\emptyset$ for $n\ge n_0$.}   (iii) Show that if
$\sum_{n=0}^{\infty}\mu E_n$ is finite, then almost every point of $X$
belongs to only finitely many $E_n$, and $\sum_{n=0}^{\infty}\mu
E_n=\sum_{k=0}^{\infty}k\mu G_k$, where

\Centerline{$G_k=H_k\setminus H_{k+1}=\{x:\#(\{n:x\in E_n\})=k\}$.}

\spheader 112Yf Let $X$ be a set and $\mu$, $\nu$ two measures on $X$, with
domains $\Sigma$, $\Tau$ respectively.   Set $\Lambda=\Sigma\cap\Tau$ and
define $\lambda:\Lambda\to[0,\infty]$ by setting $\lambda E=\mu E+\nu E$
for every $E\in\Lambda$.   Show that $(X,\Lambda,\lambda)$ is a measure
space.
%+
}%end of exercises

\endnotes{
\Notesheader{112} The calculations in such results as 112Ca-112Cc,
112Xa and 112Xc, involving only finitely many sets, are common to any
additive concept of measure;  you may have encountered them in
elementary probability theory, but of course I am now asking you to
consider also the possibility that one or more of the sets has measure
$\infty$.   I hope you will find that these results are entirely natural
and unsurprising.    I recommend Venn diagrams in this context;   a
result of this kind involving only finitely many measurable sets and
only {\it addition}, with no {\it subtraction}, will be valid in general
if and only if it is valid for the area of simple geometric shapes in
the plane.   The requirement `$\mu E<\infty$' in 112Xc is necessary
only because we are subtracting $\mu E$;   the corresponding additive
result $\mu(F\symmdiff E)+\mu E=\mu F+2\mu(E\setminus F)$ is true for
all measurable $E$ and $F$.
Of course when sequences of sets enter the picture, we need to take a
bit more care;  the results 112Cd-112Cf are the basic ones to learn.   I
think however that the only trap is in the condition `some $\mu E_n$
is finite' in 112Cf.  As noted in the remark at the end of 112C, this is
essential, and for a decreasing sequence of measurable sets it is
possible for the measure of the limit to be strictly less than the limit
of the measures, though only when the latter is infinite.
}%end of notes

\discrpage

\frfilename{mt113.tex}
\versiondate{6.4.05}
\copyrightdate{1999}

\def\chaptername{Measure spaces}
\def\sectionname{Outer measures and \Caratheodory's construction}

\newsection{113}

I introduce the most important method of constructing measures.

\leader{113A}{Outer \dvrocolon{measures}}\cmmnt{ I come now to the
third basic definition of this chapter.

\medskip

\noindent}{\bf Definition} Let $X$ be a set.   An {\bf outer measure} on
$X$ is a function $\theta:\Cal PX\to[0,\infty]$ such that

\qquad(i) $\theta \emptyset=0$,

\qquad(ii) if $A\subseteq B\subseteq X$ then $\theta A\le\theta B$,

\qquad(iii) for every sequence $\langle A_n\rangle_{n\in\Bbb N}$ of
subsets of $X$,
$\theta(\bigcup_{n\in\Bbb N}A_n)\le\sum_{n=0}^{\infty}\theta A_n$.

\cmmnt{
\leader{113B}{Remarks (a)} For comments on the use of
`$\infty$', see 112B.

\header{113Bb}{\bf (b)} Yet again, the most important outer measures
must wait until
\S\S114-115.   The idea of the `outer' measure of a set $A$ is
that it should be some kind of upper bound for the possible measure of
$A$.   If we are lucky, it may actually be the measure of $A$;  but this
is likely to be true only for sets with adequately smooth boundaries.

\header{113Bc}{\bf (c)} Putting (i) and (iii) of the definition
together, we see that
if $\theta$ is an outer measure on $X$, and $A$, $B$ are two subsets of
$X$, then $\theta(A\cup B)\le\theta A+\theta B$;  compare 112Ca and 112Cc.
}%end of comment

\leader{113C}{\Caratheodory's Method:  Theorem} Let $X$ be a set and
$\theta$ an outer measure on $X$.   Set

\Centerline{$\Sigma=\{E:E\subseteq X,\,
\theta A=\theta(A\cap E)+\theta(A\setminus E)$ for every $A\subseteq X\}$.}

\noindent Then $\Sigma$ is a $\sigma$-algebra of subsets of $X$.
Define $\mu:\Sigma\to[0,\infty]$ by writing $\mu E=\theta E$ for
$E\in\Sigma$;  then $(X,\Sigma,\mu)$ is a measure space.

\proof{{\bf (a)} The first step is to note that for any $E$,
$A\subseteq X$ we have
$\theta(A\cap E)+\theta(A\setminus E)\ge\theta A$, by 113Bc;  so that

\Centerline{$\Sigma=\{E:E\subseteq X,\,
\theta A\ge\theta(A\cap E)+\theta(A\setminus E)$
for every $A\subseteq X\}$.}


\medskip

{\bf (b)} Evidently $\emptyset\in\Sigma$, because

\Centerline{$\theta(A\cap\emptyset)+\theta(A\setminus\emptyset)
=\theta\emptyset+\theta A=\theta A$}

\noindent for every $A\subseteq X$.   If $E\in\Sigma$, then
$X\setminus E\in\Sigma$, because

\Centerline{$\theta(A\cap(X\setminus E))+\theta(A\setminus(X\setminus E))
=\theta(A\setminus E)+\theta(A\cap E)=\theta A$}

\noindent for every $A\subseteq X$.

\medskip

{\bf (c)} Now suppose that $E$, $F\in\Sigma$ and $A\subseteq X$.   Then

\ifdim\pagewidth<391pt
\startsideshiftedpicture
\hskip-20pt
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\sideshiftedpicture{mt113c2}{-15pt}{90.05pt}{100.68pt}
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$$\eqalignno{\theta(A\cap(&E\cup F))+\theta(A\setminus(E\cup F))
    &\text{diagram (i)}\cr
&=\theta(A\cap(E\cup F)\cap E)+\theta(A\cap(E\cup F)\setminus E)
  +\theta(A\setminus(E\cup F))
    &\text{diag.\ (ii)}\cr
\noalign{\noindent (because $E\in\Sigma$ and
$A\cap(E\cup F)\subseteq X)$}
&=\theta(A\cap E)+\theta((A\setminus E)\cap F)
  +\theta((A\setminus E)\setminus F)\cr
&=\theta(A\cap E)+\theta(A\setminus E)
    &\text{diag.\ (iii)}\cr
\noalign{\noindent (because $F\in\Sigma$)}
&=\theta A
    &\text{diag.\ (iv)}\cr}$$

\noindent (again because $E\in\Sigma$).   Because $A$ is arbitrary,
$E\cup F\in\Sigma$.

\medskip

{\bf (d)} Thus $\Sigma$ is closed under simple unions and complements,
and contains $\emptyset$.
%Observe that it follows that
%$E\setminus F=X\setminus(F\cup(X\setminus E))$ belongs to $\Sigma$ whenever $E$,
%$F\in\Sigma$.
Now suppose that $\langle E_n\rangle_{n\in\Bbb N}$ is a
sequence in $\Sigma$, with $E=\bigcup_{n\in\Bbb N}E_n$.   Set

\Centerline{$G_n=\bigcup_{m\le n}E_m$;}

\noindent then $G_n\in\Sigma$ for each $n$, by induction on $n$.   Set

\Centerline{$F_0=G_0=E_0$,\quad
$F_n=G_n\setminus G_{n-1}=E_n\setminus G_{n-1}$ for $n\ge 1$;}

\noindent  then
%every $F_n$ belongs to $\Sigma$, and
$E=\bigcup_{n\in\Bbb N}F_n=\bigcup_{n\in\Bbb N}G_n$.

Take any $n\ge 1$ and any $A\subseteq X$.   Then

$$\eqalign{\theta(A\cap G_n)
&=\theta(A\cap G_n\cap G_{n-1})+\theta(A\cap G_n\setminus G_{n-1})\cr
&=\theta(A\cap G_{n-1})+\theta(A\cap F_n).\cr}$$

\noindent An induction on $n$ shows that $\theta(A\cap
G_n)=\sum_{m=0}^n\theta(A\cap F_m)$ for every $n\ge 0$.

Suppose that $A\subseteq X$.   Then $A\cap
E=\bigcup_{n\in\Bbb N}A\cap F_n$, so

$$\eqalign{\theta(A\cap E)
&\le\sum_{n=0}^{\infty}\theta(A\cap F_n)\cr
&=\lim_{n\to\infty}\sum_{m=0}^n\theta(A\cap F_m)
=\lim_{n\to\infty}\theta(A\cap G_n).}$$

\noindent On the other hand,

$$\eqalign{\theta(A\setminus E)
&=\theta(A\setminus\bigcup_{n\in\Bbb N}G_n)\cr
&\le\inf_{n\in\Bbb N}\theta(A\setminus G_n)
=\lim_{n\to\infty}\theta(A\setminus G_n),}$$

\noindent using 113A(ii) to see that
$\langle\theta(A\setminus G_n)\rangle_{n\in\Bbb N}$ is non-increasing
and that $\theta(A\setminus E)\le \theta(A\setminus G_n)$ for every $n$.
Accordingly

$$\eqalign{\theta(A\cap E)+\theta(A\setminus E)
&\le\lim_{n\to\infty}\theta(A\cap G_n)
+\lim_{n\to\infty}\theta(A\setminus G_n)\cr
&=\lim_{n\to\infty}(\theta(A\cap G_n)+\theta(A\setminus G_n))
=\theta A\cr}$$

\noindent because every $G_n$ belongs to $\Sigma$, so $\theta(A\cap
G_n)+\theta(A\setminus G_n)=\theta A$ for every $n$.   But $A$ is
arbitrary, so $E\in\Sigma$, by the remark in (a) above.

Because $\langle E_n\rangle_{n\in\Bbb N}$ is arbitrary,
condition (iii) of 111A is satisfied, and $\Sigma$ is a $\sigma$-algebra
of subsets of $X$.

\medskip

{\bf (e)} Now let us turn to $\mu$, the restriction of $\theta$ to
$\Sigma$, and
Definition 112A.   Of course $\mu\emptyset=\theta\emptyset=0$.   So let
$\langle E_n\rangle_{n\in\Bbb N}$ be any disjoint sequence in $\Sigma$.
Set $G_n=\bigcup_{m\le n}E_m$ for each $n$, as in (d), and

\Centerline{$E=\bigcup_{n\in\Bbb N}E_n=\bigcup_{n\in\Bbb N}G_n$.}

\noindent As in (d),

$$\eqalign{\mu G_{n+1}
=\theta G_{n+1}
&=\theta(G_{n+1}\cap E_{n+1})+\theta(G_{n+1}\setminus E_{n+1})\cr
&=\theta E_{n+1}+\theta G_n
=\mu E_{n+1}+\mu G_n\cr}$$

\noindent for each $n$, so $\mu G_n=\sum_{m=0}^n\mu E_m$ for every $n$.

Now

\Centerline{$\mu E
=\theta E
\le\sum_{n=0}^{\infty}\theta E_n
=\sum_{n=0}^{\infty}\mu E_n$.}

\noindent But also

\Centerline{$\mu E=\theta E\ge\theta G_n=\mu G_n=\sum_{m=0}^n\mu E_m$}

\noindent for each $n$, so $\mu E\ge\sum_{n=0}^{\infty}\mu E_n$.

Accordingly $\mu E=\sum_{n=0}^{\infty}\mu E_n$.   As
$\langle E_n\rangle_{n\in\Bbb N}$ is arbitrary, 112A(iii-$\beta$) is
satisfied and $(X,\Sigma,\mu)$ is a measure space.
}%end of proof of 113C

\leader{113D}{Remark} Note\cmmnt{ from (a) in the proof above} that in
this construction

\cmmnt{\Centerline{$\Sigma=\{E:E\subseteq X,\,\theta(A\cap
E)+\theta(A\setminus
E)\le\theta A$ for every $A\subseteq X\}$\dvro{,}{.}}

\noindent Since $\theta(A\cap E)+\theta(A\setminus E)$ is
necessarily less than or equal to $\theta A$ when $\theta A=\infty$,}

\Centerline{$\Sigma=\{E:E\subseteq X,\,\theta(A\cap E)+\theta(A\setminus
E)\le\theta A$ whenever $A\subseteq X$ and $\theta A<\infty\}$.}

\exercises{
\leader{113X}{Basic exercises $\pmb{>}$(a)}
%\spheader 113Xa
Let $X$ be a set and $\theta$ an outer measure on $X$, and let
$\mu$ be the measure on $X$ defined from $\theta$ by \Caratheodory's
method.   Show that if $\theta A=0$, then $\mu$ measures $A$, so that a
set $A\subseteq X$ is $\mu$-negligible iff $\theta A=0$, and $\mu$ is `complete' in
the sense of 112Df.

\spheader 113Xb Let $X$ be a set.
{(i)} Show that if $\theta_1$, $\theta_2$ are outer measures on $X$, so is
$\theta_1+\theta_2$, setting
$(\theta_1+\theta_2)(A)=\theta_1A+\theta_2A$ for every $A\subseteq X$.
{(ii)} Show that if
$\langle\theta_i\rangle_{i\in I}$ is any non-empty family of
outer measures on $X$, so is $\theta=\sup_{i\in I}\theta_i$, setting
$\theta A=\sup_{i\in I}\theta_iA$ for every $A\subseteq X$.   {(iii)}
Show that if $\theta_1$, $\theta_2$ are outer measures on $X$ so is
$\theta_1\wedge\theta_2$, setting

\Centerline{$(\theta_1\wedge\theta_2)(A)
=\inf\{\theta_1B+\theta_2(A\setminus B):B\subseteq A\}$}

\noindent for every $A\subseteq X$.

\sqheader 113Xc Let $X$ and $Y$ be sets, $\theta$ an outer
measure on $X$, and $f:X\to Y$ a function.   Show that the functional
$B\mapsto\theta(f^{-1}[B]):\Cal PY\to[0,\infty]$ is an outer measure on
$Y$.

\sqheader 113Xd Let $X$ be a set and $\theta$ an outer measure
on $X$;  let $Y$ be any subset of $X$.   {(i)} Show that
$\theta\restrp\Cal PY$, the restriction of $\theta$ to subsets of $Y$, is
an outer measure on $Y$.   {(ii)} Show that if $E\subseteq X$ is
measured by the measure on $X$ defined from $\theta$ by
\Caratheodory's method, then $E\cap Y$ is measured by the measure on
$Y$ defined from $\theta\restrp\Cal PY$.

\sqheader 113Xe Let $X$ and $Y$ be sets, $\theta$ an outer
measure on $Y$, and $f:X\to Y$ a function.   Show that the functional
$A\mapsto\theta(f[A]):\Cal PX\to[0,\infty]$ is an outer measure.

\spheader 113Xf Let $X$ and $Y$ be sets, $\theta$ an outer
measure on $X$, and $R\subseteq X\times Y$ a relation.   Show that
the map $B\mapsto\theta(R^{-1}[B]):\Cal PY\to[0,\infty]$ is an outer
measure on $Y$, where $R^{-1}[B]=\{x:\exists\,y\in B,\,(x,y)\in R\}$
(1A1Bc).   Explain how this is a common generalization of (c),
(d-i) and (e) above, and how it can be proved by putting them together.

\spheader 113Xg Let $X$ be a set and $\theta$ an outer measure
on $X$.   Suppose that $E\subseteq X$ is measured by the measure on
$X$ defined from $\theta$ by \Caratheodory's method.   Show that
$\theta(E\cap A)+\theta(E\cup A)=\theta E+\theta A$ for every
$A\subseteq X$.

\spheader 113Xh Let $X$ be a set and $\theta:\Cal PX\to[0,\infty]$ a functional such that $\theta\emptyset=0$, $\theta A\le\theta B$ whenever $A\subseteq B\subseteq X$, and $\theta(A\cup B)\le\theta A+\theta B$ whenever $A$, $B\subseteq X$.   Set

\Centerline{$\Sigma=\{E:E\subseteq X,\,
\theta A=\theta(A\cap E)+\theta(A\setminus E)$ for every
$A\subseteq X\}$.}

\noindent Show that $\emptyset$, $X\setminus E$ and $E\cup F$ belong to
$\Sigma$ for all $E$,
$F\in\Sigma$, so that $E\setminus F$, $E\cap F\in\Sigma$ for all $E$,
$F\in\Sigma$.   Show that $\theta(E\cup F)=\theta E+\theta F$ whenever
$E$, $F\in\Sigma$ and $E\cap F=\emptyset$.

\leader{113Y}{Further exercises (a)}
%\spheader 113Ya
Let $(X,\Sigma,\mu)$ be a measure space.   For $A\subseteq X$
set $\mu^*A=\inf\{\mu E:E\in\Sigma,\,A\subseteq E\}$.   Show that for
every $A\subseteq X$ the infimum is attained, that is, there is an
$E\in\Sigma$ such that $A\subseteq E$ and $\mu E=\mu^*A$.   Show that
$\mu^*$ is an outer measure on $X$.

\spheader 113Yb Let $(X,\Sigma,\mu)$ be a measure space and $D$
any subset of
$X$.   Show that $\Sigma_D=\{E\cap D:E\in\Sigma\}$ is a $\sigma$-algebra
of subsets of $D$.   Set $\mu_D=\mu^*\restr \Sigma_D$, the function with
domain $\Sigma_D$ such that $\mu_DB=\mu^*B$ for every $B\in\Sigma_D$,
where $\mu^*$ is defined as in (a) above;  show that
$(D,\Sigma_D,\mu_D)$ is a measure space.   ($\mu_D$ is the {\bf subspace
measure} on $D$.)

\spheader 113Yc Let $(X,\Sigma,\mu)$ be a measure space and let
$\mu^*$ be the
associated outer measure on $X$, as in 113Ya.   Let $\check\mu$ be the
measure on $X$ constructed by \Caratheodory's method from $\mu^*$, and
$\check\Sigma$ its domain.   Show that $\Sigma\subseteq\check\Sigma$ and
that $\check\mu$ extends $\mu$.

\spheader 113Yd Let $X$ be a set and
$\tau:\Cal PX\to[0,\infty]$ any function such that $\tau\emptyset=0$.
For $A\subseteq X$ set

$$\eqalign{\theta A
=\inf\{\sum_{j=0}\sp{\infty}\tau C_j:\langle C_j\rangle_{j\in\Bbb N}
\text{ is a sequence of }&\text{subsets of }X\cr
&\text{such that }
A\subseteq\bigcup_{j\in\Bbb N}C_j\}.\cr}$$

\noindent Show that $\theta$ is an outer measure on $X$.   \Hint{you
will need 111F(b-ii) or something equivalent.}

\spheader 113Ye Let $X$ be a set and $\theta_1$, $\theta_2$ two
outer measures on $X$.   Show that $\theta_1\wedge\theta_2$, as
described in 113Xb(iii), is the outer measure derived by the process of
113Yd from the functional $\tau C=\min(\theta_1C,\theta_2C)$.

\spheader 113Yf Let $X$ be a set and
$\langle\theta_i\rangle_{i\in I}$ any non-empty family of outer measures
on $X$.   Set $\tau C=\inf_{i\in I}\theta_iC$ for each 
$C\subseteq X$.   Show that the outer measure derived from $\tau$ by the
process of 113Yd is the largest outer measure $\theta$ such that
$\theta A\le\theta_iA$ whenever $A\subseteq X$ and $i\in I$.

\spheader 113Yg Let $X$ be a set and $\phi:\Cal PX\to[0,\infty]$
a functional such that

\inset{$\phi\emptyset=0$;

$\phi(A\cup B)\ge\phi A+\phi B$ for all disjoint $A$, $B\subseteq X$;

if $\sequencen{A_n}$ is a non-increasing sequence of subsets of
$X$ and $\phi A_0<\infty$ then
$\phi(\bigcap_{n\in\Bbb N}A_n)=\lim_{n\to\infty}\phi A_n$;

if $\phi A=\infty$ and $a\in\Bbb R$ there is a $B\subseteq A$ such
that $a\le\phi B<\infty$.}

\noindent Set

\Centerline{$\Sigma=\{E:E\subseteq X,\,\phi(A\cap E)+\phi(A\setminus
E)=\phi A$ for every $A\subseteq X\}$.}

\noindent Show that $(X,\Sigma,\phi\restr\Sigma)$ is a measure space.

\spheader 113Yh Let $(X,\Sigma,\mu)$ be a measure space and for
$A\subseteq X$ set
$\mu_*A=\sup\{\mu E:E\in\Sigma$, $E\subseteq A$, $\mu E<\infty\}$.
Show that $\mu_*:\Cal PX\to[0,\infty]$ satisfies the
conditions of
113Yg, and that if $\mu X<\infty$ then the measure defined from $\mu_*$
by the method of 113Yg extends $\mu$.

\spheader 113Yi Let $X$ be a set and $\Cal A$ an {\bf algebra}
of subsets of $X$, that is, a family of subsets of $X$ such that

\inset{$\emptyset\in\Cal A$,

$X\setminus E\in\Cal A$ for every $E\in\Cal A$,

$E\cup F\in\Cal A$ whenever $E$, $F\in\Cal A$.}

\noindent Let $\phi:\Cal A\to[0,\infty]$ be a function such that

\inset{$\phi\emptyset=0$,

$\phi(E\cup F)=\phi E+\phi F$ whenever $E$, $F\in\Cal A$
and $E\cap F=\emptyset$,

$\phi E=\lim_{n\to\infty}\phi E_n$ whenever
$\sequencen{E_n}$ is a non-decreasing sequence in $\Cal A$ with union
$E$.}

\noindent Show that there is a measure $\mu$ on $X$ extending $\phi$.
\Hint{set $\phi A=\infty$ for $A\in\Cal PX\setminus\Cal A$;
define $\theta$ from $\phi$ as in 113Yd, and $\mu$ from $\theta$.}

\spheader 113Yj (T.de Pauw) Let $X$ be a set, $\Tau$ a
$\sigma$-algebra of subsets of $X$, and $\theta$ an outer measure on
$X$.   Set
$\Sigma=\{E:E\in\Tau,\,\theta A=\theta(A\cap E)+\theta(A\setminus E)$
for every $A\in\Tau\}$.   Show that $\Sigma$ is
a $\sigma$-algebra of subsets of $X$ and that $\theta\restr\Sigma$ is a
measure.

\spheader 113Yk\dvAnew{2010} Let $X$, $\tau:\Cal PX\to[0,\infty]$ and
$\theta$ be as in 113Yd;  let $\mu$ be the measure defined by
\Caratheodory's method from $\theta$, and $\Sigma$ the domain of $\mu$.  
Suppose that $E\subseteq X$ is such that 
$\theta(C\cap E)+\theta(C\setminus E)\le\tau C$ whenever $C\subseteq X$
is such that $0<\tau C<\infty$.   Show that $E\in\Sigma$.
}%end of exercises

\endnotes{
\Notesheader{113} We are proceeding by the easiest stages I can devise
to the construction of a non-trivial measure space, that is, Lebesgue
measure on $\Bbb R$.   There are many constructions of Lebesgue measure,
but in my view \Caratheodory's method (113C) is the right one to begin
with, because it is the most powerful and versatile single technique for
constructing measures.   It is, of course, abstract -- it deals with
arbitrary outer measures on arbitrary sets;  but I really think that the
Lebesgue theory, intertwined as it is with the rich structure of Euclidean
space, is harder than the abstract theory of measure.
We do at least have here a serious theorem for you to get your
teeth into, mastery of which should be both satisfying and useful.   I
must say that I think it very remarkable that such a direct construction
should be effective.    Looking at the proof, it is perhaps worth while
distinguishing between the `algebraic' or `finite' parts ((a)-(c)) and
the parts involving sequences of sets ((d)-(e));   the former amount to
a proof of 113Xh.   Outer measures of various kinds appear throughout
measure theory, and I sketch a few of the relevant constructions in
113X-113Y.
}%end of notes


\discrpage

\frfilename{mt114.tex}
\versiondate{14.6.05}
\copyrightdate{1994}

\def\chaptername{Measure spaces}
\def\sectionname{Lebesgue measure on $\Bbb R$}

\newsection{114}
\def\headlinesectionname{Lebesgue measure on ${\eightBbb R}$}

Following the very abstract ideas of \S\S111-113, we have an urgent
need for a non-trivial example of a measure space.   By far the most
important example is the real line with Lebesgue measure, and I now
proceed to a description of this measure (114A-114E), with a few of its
basic properties.

The principal ideas of this section are repeated in \S115, and if
you have encountered Lebesgue measure before, or feel confident in your
ability to deal with two- and three-dimensional spaces at the same time
as doing some difficult analysis, you could go directly to that
section, turning back to this one only when a specific reference is
given.

\leader{114A}{Definitions (a)} For the purposes of this section, a {\bf
half-open interval} in $\Bbb R$ is a set of the form
$\coint{a,b}=\{x:a\le x<b\}$, where $a$, $b\in\Bbb R$.

\cmmnt{Observe that I allow $b\le a$ in this formula;  in this case
$\coint{a,b}=\emptyset$ (see 1A1A).}

\header{114Ab}{\bf (b)}\cmmnt{ If $I\subseteq \Bbb R$ is a half-open
interval, then either
$I=\emptyset$ or $I=\coint{\inf I,\sup I}$, so that its endpoints are
well defined.}   We\cmmnt{ may therefore} define the {\bf length}
$\lambda I$ of a half-open interval $I$ by setting

\Centerline{$\lambda\emptyset = 0$,
\quad $\lambda\coint{a,b}=b-a$ if $a<b$.}

\leader{114B}{Lemma} If $I\subseteq\Bbb R$ is a half-open interval and
$\langle I_j\rangle_{j\in\Bbb N}$ is a sequence of half-open intervals
covering $I$, then $\lambda I\le\sum_{j=0}^{\infty}\lambda I_j$.

\proof{{\bf (a)}
If $I=\emptyset$ then of course $\lambda I=0\le\sum_{j=0}^{\infty}\lambda
I_j$.   Otherwise, take
$I=\coint{a,b}$, where $a<b$.    For each $x\in \Bbb R$ let $H_x$ be the
half-line $\ooint{-\infty,x}$, and consider the set

\Centerline{$A=\{x:a\le x\le b,\,
x-a\le\sum_{j=0}\sp{\infty}\lambda(I_j\cap
H_x)\}$.}

\noindent (Note that if $I_j=\coint{c_j,d_j}$ then $I_j\cap
H_x=\coint{c_j,\min(d_j,x)}$, so $\lambda(I_j\cap H_x)$ is always
defined.)   We have $a\in A$ (because
$a-a=0\le\sum_{j=0}\sp{\infty}\lambda(I_j\cap
H_a)$) and of course $A\subseteq[a,b]$, so $c=\sup A$ is defined, and
belongs to $[a,b]$.

\medskip

{\bf (b)} We find now that $c\in A$.

\prooflet{

$$\eqalign{\Prf\ c-a&=\sup_{x\in A}x-a\cr
&\le\sup_{x\in A}\sum_{j=0}^{\infty}\lambda(I_j\cap H_x)
\le\sum_{j=0}^{\infty}\lambda(I_j\cap H_c). \text{ \Qed}\cr}$$
}

\medskip

{\bf (c)} \Quer\ Suppose, if possible, that $c<b$.   Then $c\in
\coint{a,b}$, so there is some $k\in\Bbb N$ such that $c\in I_k$.
Express $I_k$ as $\coint{c_k,d_k}$;  then $x=\min(d_k,b)>c$.    For each
$j$, $\lambda(I_j\cap H_x)\ge\lambda(I_j\cap H_c)$, while

\Centerline{$\lambda(I_k\cap H_x)=\lambda(I_k\cap H_c)+x-c$.}

\noindent So

$$\eqalign{\sum_{j=0}^{\infty}\lambda(I_j\cap H_x)
&\ge\sum_{j=0}^{\infty}\lambda(I_j\cap H_c)+x-c\cr
&\ge c-a+x-c
=x-a,\cr}$$

\noindent so $x\in A$;  but $x>c$ and $c=\sup A$.  \Bang

\medskip

{\bf (d)} We conclude that $c=b$, so that $b\in A$ and

\Centerline{$b-a\le\sum_{j=0}^{\infty}\lambda(I_j\cap
H_b)\le\sum_{j=0}^{\infty}\lambda I_j$,}

\noindent as claimed.
}%end of proof of 114B


\vleader{36pt}{114C}{Definition} Now\cmmnt{, and for the rest of this
section,} define $\theta:\Cal P\Bbb R\to[0,\infty]$ by
writing

$$\eqalign{\theta A=\inf\{\sum_{j=0}\sp{\infty}\lambda I_j:\langle
I_j\rangle_{j\in\Bbb N}
\text{ is a sequence of }&\text{half-open intervals}\cr
&\text{such that }
A\subseteq\bigcup_{j\in\Bbb N}I_j\}.\cr}$$

\noindent\cmmnt{Observe that every $A$ can be covered by some sequence
of half-open intervals -- e.g.,
$A\subseteq\bigcup_{n\in\Bbb N}\coint{-n,n}$;  so that if we interpret
the sums in $[0,\infty]$, as in 112Bc above, we always have a non-empty
set to take the infimum of, and $\theta A$ is always defined in
$[0,\infty]$.
This function} $\theta$ is called {\bf Lebesgue outer measure} on
$\Bbb R$\cmmnt{;  the phrase is justified by (a) of the next proposition}.



\leader{114D}{Proposition} (a) $\theta$ is an outer measure on $\Bbb R$.

(b) $\theta I=\lambda I$ for every half-open interval
$I\subseteq\Bbb R$.

\proof{{\bf (a)(i)} $\theta$ takes values in $[0,\infty]$
because every $\theta A$ is the infimum of a non-empty subset of
$[0,\infty]$.

\medskip

\quad{\bf (ii)} $\theta\emptyset=0$ because (for instance) if we set
$I_j=\emptyset$ for every $j$, then every $I_j$ is a half-open interval
(on the convention I am using) and
$\emptyset\subseteq\bigcup_{j\in\Bbb N}I_j$, $\sum_{j=0}\sp{\infty}\lambda I_j=0$.

\medskip

\quad{\bf (iii)} If $A\subseteq B$ then whenever
$B\subseteq\bigcup_{j\in\Bbb N}I_j$ we have
$A\subseteq\bigcup_{j\in\Bbb N}I_j$, so $\theta A$ is the infimum of a set at least as large as that
involved in the definition of $\theta B$, and $\theta A\le\theta B$.

\medskip

\quad{\bf (iv)} Now suppose that $\langle A_n\rangle_{n\in\Bbb N}$ is a
sequence of subsets of $\Bbb R$, with union $A$.   For any $\epsilon>0$,
we can choose, for each $n\in\Bbb N$, a sequence $\langle
I_{nj}\rangle_{j\in\Bbb N}$ of half-open intervals such that
$A_n\subseteq\bigcup_{j\in\Bbb N}I_{nj}$ and
$\sum_{j=0}\sp{\infty}\lambda I_{nj}\le\theta A_n+2^{-n}\epsilon$.
(You should perhaps check that this formulation is valid whether $\theta
A_n$ is finite or infinite.)   Now by 111F(b-ii)
there is a bijection from $\Bbb N$ to $\Bbb N\times\Bbb N$;  express
this in the form $m\mapsto(k_m,l_m)$.   Then $\sequence{m}{I_{k_m,l_m}}$
is a sequence of half-open intervals, and

\Centerline{$A\subseteq\bigcup_{m\in\Bbb N}I_{k_m,l_m}$.}

\noindent \Prf\  If $x\in A=\bigcup_{n\in\Bbb N}A_n$ there must be an
$n\in\Bbb N$ such that $x\in A_n\subseteq\bigcup_{j\in\Bbb N}I_{nj}$, so
there is a $j\in\Bbb N$ such that $x\in I_{nj}$.   Now
$m\mapsto(k_m,l_m)$ is surjective, so there is an $m\in\Bbb N$ such that
$k_m=n$ and $l_m=j$, in which case $x\in I_{k_m,l_m}$.\ \Qed

Next,

\Centerline{$\sum_{m=0}\sp{\infty}\lambda I_{k_m,l_m}
\le\sum_{n=0}\sp{\infty}\sum_{j=0}\sp{\infty}\lambda I_{nj}$.}

\noindent \Prf\ If $M\in\Bbb N$, then
$N=\max(k_0,k_1,\ldots,k_M,l_0,l_1,\ldots,l_M)$ is finite;   because
every $\lambda I_{nj}$ is
greater than or equal to $0$, and any pair $(n,j)$ can appear at most
once as a $(k_m,l_m)$,

\Centerline{$\sum_{m=0}^M\lambda I_{k_m,l_m}
\le\sum_{n=0}^N\sum_{j=0}^N\lambda I_{nj}
\le\sum_{n=0}^N\sum_{j=0}^{\infty}\lambda I_{nj}
\le\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}\lambda I_{nj}$.}

\noindent So

\Centerline{$\sum_{m=0}^{\infty}\lambda I_{k_m,l_m}
=\lim_{M\to\infty}\sum_{m=0}^M\lambda I_{k_m,l_m}
\le\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}\lambda I_{nj}$.  \Qed}

Accordingly

$$\eqalign{\theta A
&\le\sum_{m=0}\sp{\infty}\lambda I_{k_m,l_m}\cr
&\le\sum_{n=0}\sp{\infty}\sum_{j=0}\sp{\infty}\lambda I_{nj}\cr
&\le\sum_{n=0}\sp{\infty}(\theta A_n+2^{-n}\epsilon)\cr
&=\sum_{n=0}\sp{\infty}\theta A_n+\sum_{n=0}\sp{\infty}2^{-n}\epsilon\cr
&=\sum_{n=0}\sp{\infty}\theta A_n+2\epsilon.\cr}$$

\noindent Because $\epsilon$ is arbitrary, $\theta
A\le\sum_{n=0}\sp{\infty}\theta A_n$ (again, you should check that this
is valid whether or not $\sum_{n=0}\sp{\infty}\theta A_n$ is finite).
As $\langle A_n\rangle_{n\in\Bbb N}$ is arbitrary, $\theta$ is an outer
measure.

\medskip

{\bf (b)} Because we can always take $I_0=I$, $I_j=\emptyset$ for
$j\ge 1$, to obtain a sequence of half-open intervals covering $I$ with
$\sum_{j=0}\sp{\infty}\lambda I_j=\lambda I$, we surely have $\theta
I\le\lambda I$.   For the reverse inequality, use 114B:  if
$I\subseteq\bigcup_{j\in\Bbb N}I_j$, then $\lambda
I\le\sum_{j=0}^{\infty}\lambda I_j$;   as
$\langle I_j\rangle_{j\in\Bbb N}$ is arbitrary, $\theta I\ge\lambda I$ and $\theta I=\lambda I$, as
required.
}%end of proof of 114D

\cmmnt{\medskip

\noindent{\bf Remark} There is an ungainly shift in the argument of
(a-iv) above, in the stage

\Centerline{`$\theta A\le\sum_{m=0}^{\infty}\lambda
I_{k_m,l_m}\le\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}\lambda I_{nj}$'.}

\noindent I dare say you would have believed me if I had suppressed the
$k_m$, $l_m$ altogether and simply written `because
$A\subseteq\bigcup_{n,j\in\Bbb N}I_{nj}$, $\theta
A\le\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}\lambda I_{nj}$'.   I hope that
you will not find it too demoralizing if I suggest that such a jump is
not quite safe.   My reasons for interpolating a name for a bijection
between $\Bbb N$ and $\Bbb N\times\Bbb N$, and taking a couple of lines
to say explicitly that $\sum_{m=0}^{\infty}\lambda
I_{k_m,l_m}\le\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}\lambda I_{nj}$, are
the following.   To start with, there is the formal point that the
definition 114C demands a simple sequence, not a double sequence.   Is it
really obvious that it doesn't matter here?   If so, why?   There can be
no way to justify the shift which does not rely on the facts that
$\Bbb N\times\Bbb N$ is countable and every $\lambda I_{nj}$ is
non-negative.
If either of those were untrue, the method would be in grave danger of
failing.

At some point we shall certainly need to discuss sums over infinite index
sets other than $\Bbb N$, including uncountable index sets.   I have
already touched on these in 112Bd, and I will return to them in 226A in
Volume 2.
For the moment, I feel that we have quite enough new ideas to cope with,
and that what we need here is a reasonably honest expedient to deal with
the question immediately before us.

You may have noticed, or guessed, that some of the inequalities `$\le$'
here must actually be equalities;  if so, check your guess in 114Ya.
}%end of comment

\leader{114E}{Definition} Because Lebesgue outer
measure\cmmnt{ (114C)} is\cmmnt{ indeed} an
outer measure\cmmnt{ (114Da)},  we may use it to construct a measure
$\mu$, using \Caratheodory's method\cmmnt{ (113C)}.
This measure is {\bf Lebesgue measure on $\Bbb R$}.
The sets $E$ measured by $\mu$\cmmnt{ (that is, for
which $\theta(A\cap E)+\theta(A\setminus E)=\theta A$ for every
$A\subseteq\Bbb R$)} are called {\bf Lebesgue measurable}.

Sets which are negligible for $\mu$ are called {\bf Lebesgue
negligible}\cmmnt{;  note that these are just the sets $A$ for which
$\theta A=0$, and are all Lebesgue measurable (113Xa)}.



\leader{114F}{Lemma} Let $x\in\Bbb R$.   Then
$H_x=\ooint{-\infty,x}$ is
Lebesgue measurable for every $x\in\Bbb R$.

\proof{{\bf (a)} The point is that $\lambda I=\lambda(I\cap
H_x)+\lambda(I\setminus H_x)$ for every half-open interval
$I\subseteq\Bbb R$.   \Prf\ If either $I\subseteq H_x$ or $I\cap
H_x=\emptyset$, this is trivial.   Otherwise, $I$ must be of the form
$\coint{a,b}$, where $a<x<b$.   Now $I\cap H_x=\coint{a,x}$ and
$I\setminus H_x=\coint{x,b}$  are both half-open intervals, and

\Centerline{$\lambda(I\cap H_x)+\lambda(I\setminus H_x)=(x-a)+(b-x)
=b-a=\lambda I$. \Qed}

\medskip

{\bf (b)} Now suppose that $A$ is any subset of $\Bbb R$, and
$\epsilon>0$.   Then we can find a sequence $\langle
I_j\rangle_{j\in\Bbb N}$
of half-open intervals such that $A\subseteq\bigcup_{j\in\Bbb N}I_j$ and
$\sum_{j=0}^{\infty}\lambda I_j\le\theta A+\epsilon$.   Now
$\langle I_j\cap H_x\rangle_{j\in\Bbb N}$ and
$\langle I_j\setminus  H_x\rangle_{j\in\Bbb N}$ are sequences of
half-open intervals and
$A\cap H_x\subseteq\bigcup_{j\in\Bbb N}(I_j\cap H_x)$,
$A\setminus H_x\subseteq\bigcup_{j\in\Bbb N}(I_j\setminus H_x)$.   So

$$\eqalign{\theta(A\cap H_x)+\theta(A\setminus H_x)
&\le\sum_{j=0}^{\infty}\lambda(I_j\cap
H_x)+\sum_{j=0}^{\infty}\lambda(I_j\setminus H_x) \cr
&=\sum_{j=0}^{\infty}\lambda I_j
\le\theta A+\epsilon.\cr}$$

\noindent Because $\epsilon$ is arbitrary,
$\theta(A\cap H_x)+\theta(A\setminus H_x)\le\theta A$;
because $A$ is arbitrary,
$H_x$ is measurable, as remarked in 113D.
}%end of proof of 114F

\leader{114G}{Proposition} All Borel subsets of $\Bbb R$ are Lebesgue
measurable;  in particular, all open sets, and all sets of the following
classes, together with countable unions of them:

(i) open intervals $\left]a,b\right[$, $\left]-\infty,b\right[$,
$\left]a,\infty\right[$, $\left]-\infty,\infty\right[$, where
$a<b\in\Bbb R$;

(ii) closed intervals $[a,b]$, where $a\le b\in\Bbb R$;

(iii) half-open intervals $\left[a,b\right[$, $\left]a,b\right]$,
$\left]-\infty,b\right]$, $\left[a,\infty\right[$, where $a<b$ in
$\Bbb R$.

\noindent We have\cmmnt{ moreover} the following formula for the measures of such sets, writing $\mu$ for Lebesgue measure:

\Centerline{$\mu\left]a,b\right[=\mu[a,b]
=\mu\left[a,b\right[=\mu\left]a,b\right]=b-a$}

\noindent whenever $a\le b$ in $\Bbb R$, while all the unbounded
intervals have infinite measure.   It follows that every countable
subset of $\Bbb R$ is measurable and of zero measure.


\proof{{\bf (a)} I show first that all open subsets of $\Bbb R$
are measurable.   \Prf\ Let $G\subseteq\Bbb R$ be open.   Let
$K\subseteq\Bbb Q\times\Bbb Q$ be the set of pairs $(q,q')$ of rational
numbers such that $\coint{q,q'}\subseteq G$.   Now by the remarks in
111E-111F
-- specifically, 111Eb, showing that $\Bbb Q$ is countable, 111F(b-iii),
showing that products of countable sets are countable, and 111F(b-i),
showing that subsets of countable sets are countable -- we see that $K$
is countable.   Also, every $\coint{q,q'}$ is measurable, being
$H_{q'}\setminus H_q$ in the language of 114F.   So, by 111Fa,
$G'=\bigcup_{(q,q')\in K}\coint{q,q'}$ is measurable.

By the definition of $K$, $G'\subseteq G$.   On the other hand, if
$x\in G$, there is an $\epsilon>0$
such that $\ooint{x-\epsilon,x+\epsilon}\subseteq G$.
Now there are rational numbers $q\in\ocint{x-\epsilon,x}$ and
$q'\in\ocint{x,x+\epsilon}$, so that $(q,q')\in K$ and
$x\in\coint{q,q'}\subseteq G'$.
As $x$ is arbitrary, $G=G'$ and $G$ is measurable.\ \Qed

\medskip

{\bf (b)} Now the family $\Sigma$ of Lebesgue measurable sets is a
$\sigma$-algebra of subsets of $\Bbb R$ including the family of open
sets, so must contain every Borel set, by the definition of Borel set
(111G).
\medskip

{\bf (c)} Of the types of interval considered, all the open intervals
are actually open sets, so are surely Borel.   The complement of a
closed interval is expressible as the union of at most two open
intervals, so is Borel, and the closed interval, being the complement of
a Borel set, is Borel.   A bounded half-open interval is expressible as
the intersection of an open interval with a closed interval, so is
Borel;  and finally an unbounded interval of the form
$\left]-\infty,b\right]$ or $\left[a,\infty\right[$ is the complement of
an open interval, so is also Borel.

\medskip

{\bf (d)} To compute the measures, we already know from 114Db that

\Centerline{$\mu\coint{a,b}=\theta\coint{a,b}=b-a$}

\noindent if $a\le b$.   For the other types of bounded
interval, it is enough to note that $\mu\{a\}=0$ for every $a\in\Bbb R$,
as the different intervals differ only by one or two points;  and this
is so because $\{a\}\subseteq\coint{a,a+\epsilon}$, so
$\mu\{a\}\le\epsilon$, for every $\epsilon>0$.

As for the unbounded intervals, they include arbitrarily long half-open
intervals, so must have infinite measure.

\medskip

{\bf (e)} As just remarked, $\mu\{a\}=0$ for every $a$.   If
$A\subseteq\Bbb R$ is countable, it is either empty or expressible as
$\{a_n:n\in\Bbb N\}$.   In the former case $\mu A=\mu\emptyset=0$;  in
the latter, $A=\bigcup_{n\in\Bbb N}\{a_n\}$ is Borel and $\mu
A\le\sum_{n=0}^{\infty}\mu\{a_n\}=0$.
}%end of proof of 114G

\exercises{
\leader{114X}{Basic exercises $\pmb{>}$(a)} Let $g:\Bbb R\to\Bbb R$ be
any non-decreasing function.   For
half-open intervals $I\subseteq\Bbb R$ define $\lambda_gI$ by setting

\Centerline{$\lambda_g\emptyset=0$,\quad
$\lambda_g\coint{a,b}
=\lim_{x\uparrow b}g(x)-\lim_{x\uparrow a}g(x)$}

\noindent if $a<b$.   For any set $A\subseteq\Bbb R$ set

$$\eqalign{\theta_gA=\inf\{\sum_{j=0}\sp{\infty}\lambda_gI_j:\langle
I_j\rangle_{j\in\Bbb N}
\text{ is a sequence of }&\text{half-open intervals}\cr
&\text{such that }
A\subseteq\bigcup_{j\in\Bbb N}I_j\}.\cr}$$

\noindent Show that $\theta_g$ is an outer measure on $\Bbb R$.   Let
$\mu_g$ be the measure defined from $\theta_g$ by \Caratheodory's
method;  show that $\mu_gI$ is defined and equal to $\lambda_gI$ for every half-open interval
$I\subseteq\Bbb R$, and that every Borel subset of $\Bbb R$ is
in the domain of $\mu_g$.

($\mu_g$ is the {\bf Lebesgue-Stieltjes} measure associated with $g$.)

\spheader 114Xb At which point would the argument of 114Xa break down if
we wrote $\lambda_g\coint{a,b}=g(b)-g(a)$ instead of using the formula
given?

\sqheader 114Xc Write $\theta$ for Lebesgue outer measure and $\mu$
for Lebesgue
measure on $\Bbb R$.   Show that $\theta A=\inf\{\mu E:E$ is Lebesgue
measurable, $A\subseteq E\}$ for every $A\subseteq\Bbb R$.
\Hint{Consider sets $E$ of the form $\bigcup_{j\in\Bbb N}I_j$, where
$\sequence{j}{I_j}$ is a sequence of half-open intervals.}

\spheader 114Xd Let $X$ be a set, $\Cal I$ a family of subsets
of $X$ such
that $\emptyset\in\Cal I$, and $\lambda:\Cal I\to\coint{0,\infty}$ a
function such that $\lambda\emptyset=0$.   Define
$\theta:\Cal PX\to[0,\infty]$ by writing

\Centerline{$\theta A=\inf\{\sum_{j=0}\sp{\infty}\lambda I_j:\langle
I_j\rangle_{j\in\Bbb N}
\text{ is a sequence in }\Cal I\text{ such that }
A\subseteq\bigcup_{j\in\Bbb N}I_j\}$,}

\noindent interpreting $\inf\emptyset$ as $\infty$, so that
$\theta A=\infty$
if $A$ is not covered by any sequence in $\Cal I$.   Show that $\theta$
is an outer measure on $X$.

\spheader 114Xe Let $E\subseteq\Bbb R$ be a set of finite
measure for Lebesgue
measure $\mu$.   Show that for every $\epsilon>0$ there is a disjoint
family $I_0,\ldots,I_n$ of half-open intervals such that
$\mu(E\symmdiff\bigcup_{j\le n}I_j)\le\epsilon$.   \Hint{let
$\langle J_j\rangle_{j\in\Bbb N}$ be a sequence of half-open intervals
such that $E\subseteq\bigcup_{j\in\Bbb N}J_j$ and
$\sum_{j=0}^{\infty}\mu J_j\le\mu E+\bover{1}{2}\epsilon$.   Now take a
suitably large $m$ and express $\bigcup_{j\le m}J_j$ as a disjoint union
of half-open intervals.}

\sqheader 114Xf Write $\theta$ for Lebesgue outer measure and $\mu$
for Lebesgue
measure on $\Bbb R$.   Suppose that $c\in\Bbb R$.   Show that
$\theta(A+c)=\theta A$ for every $A\subseteq\Bbb R$, where
$A+c=\{x+c:x\in A\}$.   Show that if $E\subseteq \Bbb R$ is measurable
so is $E+c$, and that in this case $\mu(E+c)=\mu E$.

\spheader 114Xg Write $\theta$ for Lebesgue outer measure and $\mu$
for Lebesgue
measure on $\Bbb R$.   Suppose that $c>0$.   Show that
$\theta(cA)=c\theta(A)$ for every $A\subseteq\Bbb R$, where
$cA=\{cx:x\in A\}$.
Show that if $E\subseteq \Bbb R$ is measurable so is $cE$,
and that in this case $\mu(cE)=c\mu E$.

\leader{114Y}{Further exercises (a)}
%\spheader 114Ya
In (a-iv) of the proof of 114D, show that $\sum_{m=0}^{\infty}\lambda
I_{k_m,l_m}$ is actually equal to
$\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}\lambda I_{nj}$.

\spheader 114Yb Let $g$, $h:\Bbb R\to\Bbb R$ be two non-decreasing
functions, with sum $g+h$;  let $\mu_g$, $\mu_h$, $\mu_{g+h}$ be the
corresponding Lebesgue-Stieltjes measures (114Xa).   Show that

\Centerline{$\dom\mu_{g+h}=\dom\mu_g\cap\dom\mu_h$,
\quad$\mu_{g+h}E=\mu_gE+\mu_hE$ for every $E\in\dom\mu_{g+h}$.}
\spheader 114Yc Let $\sequencen{g_n}$ be a sequence of non-decreasing
functions from $\Bbb R$ to $\Bbb R$, and suppose that
$g(x)=\sum_{n=0}^{\infty}g_n(x)$ is defined and finite for every
$x\in\Bbb R$.   Let $\mu_{g_n}$, $\mu_g$ be the corresponding
Lebesgue-Stieltjes measures.   Show that

\Centerline{$\dom\mu_g=\bigcap_{n\in\Bbb N}\dom\mu_{g_n}$,
\quad$\mu_gE=\sum_{n=0}^{\infty}\mu_{g_n}E$ for every $E\in\dom\mu_g$.}

\spheader 114Yd(i) Show that if $A\subseteq\Bbb R$
and $\epsilon>0$, there is
an open set $G\supseteq A$ such that $\theta G\le\theta A+\epsilon$,
where $\theta$ is Lebesgue outer measure.  (ii) Show that if
$E\subseteq\Bbb R$ is Lebesgue measurable and $\epsilon>0$, there is an
open set $G\supseteq E$ such that $\mu(G\setminus E)\le\epsilon$, where
$\mu$ is Lebesgue measure.   \Hint{consider first the case of
bounded $E$.}   (iii) Show that if $E\subseteq\Bbb R$ is Lebesgue
measurable, there are Borel sets $H_1$, $H_2$ such that $H_1\subseteq
E\subseteq H_2$ and $\mu(H_2\setminus E)=\mu(E\setminus H_1)=0$.
\Hint{use (ii) to find $H_2$, and then consider the complement of
$E$.}

\spheader 114Ye Write $\theta$ for Lebesgue outer measure on
$\Bbb R$.   Show
that a set $E\subseteq\Bbb R$ is Lebesgue measurable iff
$\theta([-n,n]\cap E)+\theta([-n,n]\setminus E)=2n$ for every
$n\in\Bbb N$.   \Hint{Use 114Yd to show that for each $n$ there are
measurable sets $F_n$, $H_n$ such that $F_n\subseteq[-n,n]\cap
E\subseteq H_n$ and $H_n\setminus F_n$ is negligible.}

\spheader 114Yf Repeat 114Xc and 114Yd-114Ye for the
Lebesgue-Stieltjes measures of 114Xa.

\spheader 114Yg Write $\Cal B$ for the $\sigma$-algebra of Borel
subsets of
$\Bbb R$, and let $\nu:\Cal B\to[0,\infty]$ be a measure.   Let
$g$, $\lambda_g$, $\theta_g$ and $\mu_g$ be as in 114Xa.   Show that if
$\nu I=\lambda_gI$ for every half-open interval $I$, then $\nu E=\mu_gE$
for every $E\in\Cal B$.   \Hint{first consider open sets $E$, and
then use 114Yd(i) as extended in 114Yf.}

\spheader 114Yh Write $\Cal B$ for the $\sigma$-algebra of Borel
subsets of
$\Bbb R$, and let $\nu:\Cal B\to[0,\infty]$ be a measure such that
$\nu[-n,n]<\infty$ for every $n\in\Bbb N$.   Show that there is a
function $g:\Bbb R\to\Bbb R$ which is non-decreasing, continuous on the
left and
such that $\nu E=\mu_gE$ for every $E\in\Cal B$, where $\mu_g$ is
defined as in 114Xa.   Is $g$ unique?

\spheader 114Yi Write $\Cal B$ for the $\sigma$-algebra of Borel
subsets of
$\Bbb R$, and let $\nu_1$, $\nu_2$ be measures with domain $\Cal B$ such
that $\nu_1 I=\nu_2I<\infty$ for every half-open interval
$I\subseteq\Bbb R$.     Show that $\nu_1E=\nu_2E$ for every $E\in\Cal
B$.

\spheader 114Yj Let $\Cal E$ be any family of half-open
intervals in $\Bbb R$.   Show that (i) there is a countable
$\Cal C\subseteq\Cal E$ such that $\bigcup\Cal E=\bigcup\Cal C$
(definition:
1A1F) (ii) that $\bigcup\Cal E$ is a Borel set, so is Lebesgue
measurable (iii) that there is a disjoint sequence $\sequencen{I_n}$ of
half-open intervals in $\Bbb R$ such that $\bigcup\Cal
E=\bigcup_{n\in\Bbb N}I_n$.

\spheader 114Yk Show that for almost every $x\in\Bbb R$ (as measured by
Lebesgue measure) the set

\Centerline{$\{(m,n):m\in\Bbb Z,\,n\in\Bbb N\setminus\{0\},
\,|x-\Bover{m}{n}|\le\Bover{1}{n^3}\}$}

\noindent is finite.   \Hint{estimate the outer measure of
$\bigcup_{n\ge n_0}\bigcup_{|m|\le kn}
[\bover{m}{n}-\bover1{n^3},\bover{m}{n}+\bover1{n^3}]$ for $n_0$,
$k\ge 1$.}   Repeat with $2+\epsilon$ in the place of $3$.

\spheader 114Yl Write $\mu$ for Lebesgue measure on $\Bbb R$.   Show that
there is a countable family $\Cal F$ of Lebesgue measurable subsets of
$\Bbb R$ such that whenever $\mu E$ is defined and finite, and
$\epsilon>0$, there is an $F\in\Cal F$ such that
$\mu(E\symmdiff F)\le\epsilon$.   \Hint{in 114Xe, show that we can take the
$I_j$ to have rational endpoints.}
%114Xe
}%end of exercises

\endnotes{
\Notesheader{114} My own interests are in \lq abstract'
measure theory, and from the point of view of the structure of this
treatise, the chief object of this section is the description of a
non-trivial measure space to provide a focus for the general theorems
which follow.   Let me enumerate the methods of constructing measure
spaces already available to us.   (a) We have the point-supported measures of
112Bd;  in some ways, these are trivial;  but they do
occur in applications, and, just because they are generally easy to deal
with, it is often right to test any new ideas on them.   (b) We have
Lebesgue measure on $\Bbb R$;  a straightforward generalization of the
construction yields the Lebesgue-Stieltjes measures (114Xa).   (c) Next,
we have ways of building new measures from old, starting with subspace
measures (113Yb), image measures (112Xf) and sums of
measures (112Yf).   Perhaps the most important of these is
`Lebesgue measure on $[0,1]$', I call it $\mu_1$ for the moment, where
the domain of $\mu_1$ is $\{E:E\subseteq[0,1]$ is Lebesgue
measurable$\}=\{E\cap[0,1]:E\subseteq\Bbb R$ is Lebesgue measurable$\}$,
and $\mu_1E$ is just the Lebesgue measure of $E$ for each
$E\in\dom\mu_1$.   In fact the image measures of Lebesgue measure on
$[0,1]$ include a very large proportion of the probability measures
(that is, measures giving measure $1$ to the whole space) of importance
in ordinary applications.

Of course Lebesgue measure is not only the dominant guiding example for
general measure theory, but is itself the individual measure of greatest
importance for applications.   For this reason it would be
possible -- though in my view narrow-minded -- to read
chapters 12-13 of this
volume, and a substantial proportion of Volume 2, as if they applied only
to Lebesgue measure on $\Bbb R$.   This is, indeed, the context in which
most of these results were first developed.   I believe, however, that it
is often the case in mathematics, that one's understanding of a
particular construction is deepened and strengthened by an acquaintance
with related objects, and that one of the ways to an appreciation of the
nature of Lebesgue measure is through a study of its properties in the
more abstract context of general measure theory.

For any proper investigation of the applications of Lebesgue measure
theory we must wait for Volume 2.   But I include 114Yk as a hint of one
of the ways in which this theory can be used.
}%end of notes
\discrpage
\frfilename{mt115.tex}
\versiondate{21.7.05}
\copyrightdate{1994}

\def\chaptername{Measure spaces}
\def\sectionname{Lebesgue measure on $\BbbR^r$}


\newsection{115}
\def\headlinesectionname{Lebesgue measure on $\eightBbb R^r$}

Following the very abstract ideas of \S\S111-113, there is an urgent
need for non-trivial examples of measure spaces.   By far the most
important examples are the Euclidean spaces $\BbbR^r$ with Lebesgue
measure, and I now
proceed to a definition of these measures (115A-115E), with a few of
their basic properties.   Except at one point (in the proof of the
fundamental lemma 115B) this section does not rely essentially on \S114;
but nevertheless most students encountering Lebesgue measure for the
first time will find it easier to work through the one-dimensional case
carefully before embarking on the multi-dimensional case.

\leader{115A}{Definitions (a)} For practically the whole of this section
(the exception is the proof of Lemma 115B) $r$ will denote a fixed
integer greater than or equal to $1$.    I will use Roman letters $a$,
$b$, $c$, $d$, $x$, $y$ to denote members of $\BbbR^r$, and Greek letters for their coordinates, so that $a=(\alpha_1,\ldots,\alpha_r)$,
$b=(\beta_1,\ldots,\beta_r)$, $x=(\xi_1,\ldots,\xi_r)$.

\header{115Ab}{\bf (b)} For the purposes of this section, a {\bf
half-open interval} in $\BbbR^r$ is a set of the form
$\coint{a,b}=\{x:\alpha_i\le \xi_i<\beta_i\Forall i\le
r\}$, where $a$, $b\in\BbbR^r$.
Observe that I allow $\beta_i\le \alpha_i$ in this formula;  if this
happens for any $i$, then $\coint{a,b}=\emptyset$.



\header{115Ac}{\bf (c)} If $I=\coint{a,b}\subseteq \BbbR^r$ is a
half-open interval, then either
$I=\emptyset$ or

\Centerline{$\alpha_i=\inf\{\xi_i:x\in I\}$,
\quad$\beta_i=\sup\{\xi_i:x\in I\}$}

\noindent for every $i\le r$;  in the latter case, the expression of $I$
as a half-open interval is unique.
We may therefore define the {\bf $r$-dimensional volume} $\lambda I$ of
a half-open interval $I$ by setting

\Centerline{$\lambda\emptyset = 0$,
\quad $\lambda\coint{a,b}=\prod_{i=1}^r\beta_i-\alpha_i$ if
$\alpha_i<\beta_i$ for every $i$.}



\leader{115B}{Lemma} If $I\subseteq\BbbR^r$ is a half-open interval and
$\langle I_j\rangle_{j\in\Bbb N}$ is a sequence of half-open intervals
covering $I$, then $\lambda I\le\sum_{j=0}^{\infty}\lambda I_j$.

\proof{ The proof is by induction on $r$.   For this proof only,
therefore, I write $\lambda_r$ for the function defined on the half-open
intervals of $\BbbR^r$ by the formula of 115Ac.

\medskip

{\bf (a)} The argument for $r=1$, starting the induction, is similar to
the inductive step;  but rather than establish a suitable convention to
set up a trivial case $r=0$, or ask you to work out the details yourself,
I refer you to 114B, which is exactly the case $r=1$.

\medskip

{\bf (b)} For the inductive step to $r+1$, where $r\ge 1$, take a
half-open interval $I\subseteq\BbbR^{r+1}$ and $\sequence{j}{I_j}$ a
sequence of half-open intervals covering $I$.   If
$I=\emptyset$ then of course $\lambda_{r+1}
I=0\le\sum_{j=0}^{\infty}\lambda_{r+1}
I_j$.   Otherwise, express
$I$ as $\coint{a,b}$, where $\alpha_i<\beta_i$ for $i\le r+1$, and each
$I_j$ as $\coint{a^{(j)},b^{(j)}}$.   Write
$\zeta=\prod_{i=1}^r\beta_i-\alpha_i$, so that
$\lambda_{r+1}I=\zeta(\beta_{r+1}-\alpha_{r+1})$.   Fix $\epsilon>0$.
For each $\xi\in \Bbb R$ let $H_{\xi}$ be the
half-space $\{x:\xi_{r+1}<\xi\}$, and consider the set

\Centerline{$A=\{\xi:\alpha_{r+1}\le \xi\le \beta_{r+1},\,
\zeta(\xi-\alpha_{r+1})
\le(1+\epsilon)\sum_{j=0}^{\infty}\lambda_{r+1}(I_j\cap H_{\xi})\}$.}

\noindent (Note that
$I_j\cap H_{\xi}
=\hbox{$\bigl[$}a^{(j)},\tilde b^{(j)}\hbox{$\bigr[$}$,
where $\tilde\beta^{(j)}_i=\beta^{(j)}_i$ for $i\le r$ and
$\tilde\beta^{(j)}_{r+1}=\min(\beta^{(j)}_{r+1},\xi)$, so
$\lambda_{r+1}(I_j\cap H_{\xi})$ is always defined.)
We have $\alpha_{r+1}\in A$, because

\Centerline{$\zeta(\alpha_{r+1}-\alpha_{r+1})=0
\le(1+\epsilon)\sum_{j=0}^{\infty}\lambda_{r+1}(I_j\cap H_{\alpha_{r+1}})$,}

\noindent and of course $A\subseteq[\alpha_{r+1},\beta_{r+1}]$, so
$\gamma=\sup A$ is defined, and
belongs to $[\alpha_{r+1},\beta_{r+1}]$.

\medskip

{\bf (c)} We find now that $\gamma\in A$.

$$\eqalign{\Prf\ \zeta(\gamma-\alpha_{r+1})
&=\sup_{\xi\in A}\zeta(\xi-\alpha_{r+1})\cr
&\le(1+\epsilon)\sup_{\xi\in A}
  \sum_{j=0}^{\infty}\lambda_{r+1}(I_j\cap H_{\xi})
\le(1+\epsilon)\sum_{j=0}^{\infty}\lambda_{r+1}(I_j\cap H_{\gamma}).
\text{ \Qed}\cr}$$

\wheader{115B}{6}{2}{2}{12pt}

{\bf (d)} \Quer\ Suppose, if possible, that $\gamma<\beta_{r+1}$.   Then
$\gamma\in\coint{\alpha_{r+1},\beta_{r+1}}$.   Set

\Centerline{$J=\{x:x\in\BbbR^r,\,(x,\gamma)\in I\}
=\coint{a',b'}$,}

\noindent where $a'=(\alpha_1,\ldots,\alpha_r)$,
$b'=(\beta_1,\ldots,\beta_r)$, and for each $j\in\Bbb N$ set

\Centerline{$J_j=\{x:x\in\BbbR^r,\,(x,\gamma)\in I_j\}$.}

\noindent Because $I\subseteq\bigcup_{j\in\Bbb N}I_j$, we must have
$J\subseteq\bigcup_{j\in\Bbb N}J_j$.   Of course both $J$ and the $J_j$
are half-open intervals in $\BbbR^r$.    (This is one of the places
where it is helpful to count the empty set as a half-open interval.)
By the inductive hypothesis,
$\zeta=\lambda_rJ\le\sum_{j=0}^{\infty}\lambda_rJ_j$.   As $\zeta>0$,
there is an $m\in\Bbb N$ such that
$\zeta\le(1+\epsilon)\sum_{j=0}^m\lambda_rJ_j$.
Now for each $j\le m$, either $J_j=\emptyset$ or
$\alpha^{(j)}_{r+1}\le\gamma<\beta^{(j)}_{r+1}$;  set

\Centerline{$\xi=\min(\{\beta_{r+1}\}\cup\{\beta^{(j)}_{r+1}:j\le
m,\,J_j\ne\emptyset\})>\gamma$.}

\noindent Then

\Centerline{$\lambda_{r+1}(I_j\cap H_{\xi})
\ge\lambda_{r+1}(I_j\cap H_{\gamma})+(\xi-\gamma)\lambda_rJ_j$}

\noindent for every $j\le m$ such that $J_j$ is non-empty, and therefore
for every $j$.   Consequently

$$\eqalign{\zeta(\xi-\alpha_{r+1})
&=\zeta(\gamma-\alpha_{r+1})+\zeta(\xi-\gamma)\cr
&\le(1+\epsilon)\sum_{j=0}^{\infty}\lambda_{r+1}(I_j\cap H_{\gamma})
+(1+\epsilon)(\xi-\gamma)\sum_{j=0}^m\lambda_rJ_j\cr
&\le(1+\epsilon)\sum_{j=m+1}^{\infty}\lambda_{r+1}(I_j\cap H_{\gamma})
+(1+\epsilon)\sum_{j=0}^m\lambda_{r+1}(I_j\cap H_{\xi})\cr
&\le(1+\epsilon)\sum_{j=0}^{\infty}\lambda_{r+1}(I_j\cap H_{\xi}),\cr}$$

\noindent and $\xi\in A$, which is impossible.\ \Bang

\medskip

{\bf (e)} We conclude that $\gamma=\beta_{r+1}$, so that
$\beta_{r+1}\in A$ and

\Centerline{$\lambda_{r+1}I=\zeta(\beta_{r+1}-\alpha_{r+1})
\le(1+\epsilon)\sum_{j=0}^{\infty}\lambda_{r+1}(I_j\cap H_{\beta_{r+1}})
\le(1+\epsilon)\sum_{j=0}^{\infty}\lambda_{r+1}I_j$.}

\noindent As $\epsilon$ is arbitrary,

\Centerline{$\lambda_{r+1}I\le\sum_{j=0}^{\infty}\lambda_{r+1}I_j$,}

\noindent as claimed.
}%end of proof of 115B

\cmmnt{\medskip
\noindent{\bf Remark} This proof is hard work, and not everybody makes
such a
\ifUSEnglish{large}\else\fi
mouthful of it.   What is perhaps a more conventional approach is
sketched in 115Ya, using the Heine-Borel
theorem to reduce the problem to one of
finite covers, and then (very often) saying that it is trivial.   I do
not use this method, partly because we do not need the Heine-Borel
theorem elsewhere in this volume (though we shall certainly need it in
Volume 2, and I write out a proof in 2A2F), and partly because I do not
agree that the lemma is trivial when we have a finite sequence
$I_0,\ldots,I_m$ covering $I$.   I invite you to consider this for
yourself.   It seems to me that any rigorous argument must involve an
induction on the dimension, which is what I provide here.   Of course
dealing throughout with an infinite sequence makes it a little harder to
keep track of what we are doing, and I note that in fact there is a
crucial step which necessitates truncation of the sequence;  I mean the
formula

\Centerline{$\xi=\min(\{\beta_{r+1}\}\cup\{\beta^{(j)}_{r+1}:
                                          j\le m,\,J_j\ne\emptyset\})$}

\noindent in part (d) of the proof.   We certainly cannot take
$\xi=\inf\{\beta^{(j)}_{r+1}:j\in\Bbb N,\,J_j\ne\emptyset\}$, since this
is very likely to be equal to $\gamma$.   Accordingly I need some excuse
for truncating, which is in the sentence

\Centerline{As $\zeta>0$, there is an $m\in\Bbb N$ such that
$\zeta\le(1+\epsilon)\sum_{j=0}^m\lambda_rJ_j$.}

\noindent And that step is the reason for introducing the slack
$\epsilon$ into the definition of the set $A$ at the beginning of the
proof.   Apart from this modification, the structure of the argument is
supposed to reflect that of 114B;  so I hope you can use the simpler
formulae of 114B as a guide here.
}%end of comment

\leader{115C}{Definition} Now, and for the rest of this section, define
$\theta:\Cal P(\BbbR^r)\to[0,\infty]$ by
writing

$$\eqalign{\theta A=\inf\{\sum_{j=0}^{\infty}\lambda I_j:\langle
I_j\rangle_{j\in\Bbb N}
\text{ is a sequence of }&\text{half-open intervals}\cr
&\text{such that }
A\subseteq\bigcup_{j\in\Bbb N}I_j\}.\cr}$$


\cmmnt{\noindent Observe that every $A$ can be covered by some sequence
of half-open intervals -- e.g.,
$A\subseteq\bigcup_{n\in\Bbb N}\coint{-\tbf{n},\tbf{n}}$, writing
$\tbf{n}=(n,n,\ldots,n)\in\BbbR^r$;  so that if we interpret
the sums in $[0,\infty]$, as in 112Bc above, we always have a non-empty
set to take the infimum of, and $\theta A$ is always defined in
$[0,\infty]$.
}%end of comment

This function $\theta$ is called {\bf Lebesgue outer measure} on
$\BbbR^{r}$;  the phrase is justified by (a) of the next proposition.


\leader{115D}{Proposition} (a) $\theta$ is an outer measure on
$\BbbR^r$.

(b) $\theta I=\lambda I$ for every half-open interval
$I\subseteq\BbbR^r$.

\proof{{\bf (a)(i)} $\theta$ takes values in $[0,\infty]$
because every $\theta A$ is the infimum of a non-empty subset of
$[0,\infty]$.

\medskip

\quad{\bf (ii)} $\theta\emptyset=0$ because (for instance) if we set
$I_j=\emptyset$ for every $j$, then every $I_j$ is a half-open interval
(on the convention I am using),
$\emptyset\subseteq\bigcup_{j\in\Bbb N}I_j$
and $\sum_{j=0}^{\infty}\lambda I_j=0$.

\medskip

\quad{\bf (iii)} If $A\subseteq B$ then whenever
$B\subseteq\bigcup_{j\in\Bbb N}I_j$ we have
$A\subseteq\bigcup_{j\in\Bbb N}I_j$, so $\theta A$ is the infimum of a set at least as large as that
involved in the definition of $\theta B$, and $\theta A\le\theta B$.

\medskip

\quad{\bf (iv)} Now suppose that $\langle A_n\rangle_{n\in\Bbb N}$ is a
sequence of subsets of $\BbbR^r$, with union $A$.   For any
$\epsilon>0$,
we can choose, for each $n\in\Bbb N$, a sequence $\langle
I_{nj}\rangle_{j\in\Bbb N}$ of half-open intervals such that
$A_n\subseteq\bigcup_{j\in\Bbb N}I_{nj}$ and
$\sum_{j=0}^{\infty}\lambda I_{nj}\le\theta A_n+2^{-n}\epsilon$.
(You should perhaps check that this formulation is valid whether
$\theta A_n$
is finite or infinite.)   Now by 111F(b-ii) there is a
bijection from $\Bbb N$ to $\Bbb N\times\Bbb N$;  express
this in the form $m\mapsto(k_m,l_m)$.   Then we find that

\Centerline{$\sum_{m=0}^{\infty}\lambda I_{k_m,l_m}
=\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}\lambda I_{nj}$.}

\noindent (To see this, note that because every $\lambda I_{nj}$ is
greater than or equal to $0$, and $m\mapsto(k_m,l_m)$ is a bijection,
both sums are equal to

\Centerline{$\sup_{K\subseteq\Bbb N\times\Bbb N\text{ is finite}}
\sum_{(n,j)\in K}\lambda I_{nj}$.}

\noindent Or look at the argument written out in 114D.)   But now
$\langle I_{k_m,l_m}\rangle_{m\in\Bbb N}$ is a
sequence of half-open intervals and

\Centerline{$A=\bigcup_{n\in\Bbb N}A_n
\subseteq\bigcup_{n\in\Bbb N}\bigcup_{j\in\Bbb N}I_{nj}
=\bigcup_{m\in\Bbb N}I_{k_m,l_m}$,}

\noindent so

$$\eqalign{\theta A
&\le\sum_{m=0}^{\infty}\lambda I_{k_m,l_m}
=\sum_{n=0}^{\infty}\sum_{j=0}^{\infty}\lambda I_{nj}\cr
&\le\sum_{n=0}^{\infty}(\theta A_n+2^{-n}\epsilon)
=\sum_{n=0}^{\infty}\theta A_n+\sum_{n=0}^{\infty}2^{-n}\epsilon
=\sum_{n=0}^{\infty}\theta A_n+2\epsilon.\cr}$$

\noindent Because $\epsilon$ is arbitrary, $\theta
A\le\sum_{n=0}^{\infty}\theta A_n$ (again, you should check that this
is valid whether or not $\sum_{n=0}^{\infty}\theta A_n$ is finite).
As $\langle A_n\rangle_{n\in\Bbb N}$ is arbitrary, $\theta$ is an outer
measure.

\medskip

{\bf (b)} Because we can always take $I_0=I$, $I_j=\emptyset$ for
$j\ge 1$, to obtain a sequence of half-open intervals covering $I$ with
$\sum_{j=0}^{\infty}\lambda I_j=\lambda I$, we surely have $\theta
I\le\lambda I$.   For the reverse inequality, use 115B;  if
$I\subseteq\bigcup_{j\in\Bbb N}I_j$, then $\lambda
I\le\sum_{j=0}^{\infty}\lambda I_j$;   as
$\langle I_j\rangle_{j\in\Bbb N}$ is arbitrary, $\theta I\ge\lambda I$ and $\theta I=\lambda I$, as
required.
}%end of proof of 115D


\leader{115E}{Definition} Because Lebesgue outer
measure\cmmnt{ (115C)} is\cmmnt{ indeed}
an outer measure\cmmnt{ (115Da)},
we may use it to construct a measure $\mu$, using \Caratheodory's
method\cmmnt{ (113C)}.
This measure is {\bf Lebesgue measure on $\BbbR^r$}.
The sets $E$ for which $\mu E$ is defined\cmmnt{ (that is, for
which $\theta(A\cap E)+\theta(A\setminus E)=\theta A$ for every
$A\subseteq\BbbR^r$)} are called {\bf Lebesgue measurable}.

Sets which are negligible for $\mu$ are called {\bf Lebesgue
negligible}\cmmnt{;  note that these are just the sets $A$ for which
$\theta A=0$, and are all Lebesgue measurable (113Xa)}.


\leader{115F}{Lemma} If $i\le r$ and $\xi\in\Bbb R$, then
$H_{i\xi}=\{y:\eta_i<\xi\}$ is Lebesgue measurable.

\proof{ Write $H$ for $H_{i\xi}$.

\medskip

{\bf (a)} The point is that $\lambda I=\lambda(I\cap
H)+\lambda(I\setminus H)$ for every half-open interval
$I\subseteq\BbbR^r$.   \Prf\ If either $I\subseteq H$ or $I\cap
H=\emptyset$, this is trivial.   Otherwise, $I$ must be of the form
$\coint{a,b}$, where $\alpha_i<\xi<\beta_i$.   Now $I\cap H=\coint{a,x}$
and
$I\setminus H=\coint{y,b}$, where $\xi_j=\beta_j$ for $j\ne i$,
$\xi_i=\xi$, $\eta_j=\alpha_j$ for $j\ne i$, $\eta_i=\xi$, so both are
half-open intervals, and

$$\eqalign{\lambda(I\cap H)+\lambda(I\setminus H)
&=(\xi-\alpha_i)\prod_{j\ne i}(\beta_j-\alpha_j)
+(\beta_i-\xi)\prod_{j\ne i}(\beta_j-\alpha_j)\cr
&=(\beta_i-\alpha_i)\prod_{j\ne i}(\beta_j-\alpha_j)
=\lambda I.\text{ \Qed}\cr}$$

\medskip

{\bf (b)} Now suppose that $A$ is any subset of $\BbbR^r$, and
$\epsilon>0$.   Then we can find a sequence 
$\langle I_j\rangle_{j\in\Bbb N}$
of half-open intervals such that $A\subseteq\bigcup_{j\in\Bbb N}I_j$ and
$\sum_{j=0}^{\infty}\lambda I_j\le\theta A+\epsilon$.   In this case,
$\langle I_j\cap H\rangle_{j\in\Bbb N}$ and
$\langle I_j\setminus H\rangle_{j\in\Bbb N}$ are sequences of half-open
intervals, $A\cap H\subseteq\bigcup_{j\in\Bbb N}(I_j\cap H)$ and
$A\setminus H\subseteq\bigcup_{j\in\Bbb N}(I_j\setminus H)$.   So

$$\eqalign{\theta(A\cap H)+\theta(A\setminus H)
&\le\sum_{j=0}^{\infty}\lambda(I_j\cap
H)+\sum_{j=0}^{\infty}\lambda(I_j\setminus H) \cr
&=\sum_{j=0}^{\infty}\lambda I_j
\le\theta A+\epsilon.\cr}$$

\noindent Because $\epsilon$ is arbitrary,
$\theta(A\cap H)+\theta(A\setminus H)\le\theta A$;
because $A$ is arbitrary, $H$ is measurable, as remarked in 113D.
}%end of proof of 115F

\leader{115G}{Proposition}\dvArevised{2015} 
All Borel subsets of $\BbbR^r$ are Lebesgue
measurable;  in particular, all open sets, and all sets of the following
classes, together with countable unions of them:

\inset{open intervals
$\ooint{a,b}=\{x:x\in\BbbR^r,\,
\alpha_i<\xi_i<\beta_i\Forall i\le r\}$,
where $\alpha_i$, $\beta_i\in\Bbb R\cup\{-\infty,\infty\}$ 
for each $i\le r$;

closed intervals $[a,b]
=\{x:x\in\BbbR^r,\,\alpha_i\le\xi_i\le\beta_i
  \Forall i\le r\}$, 
where $\alpha_i$, $\beta_i\in\Bbb R\cup\{-\infty,\infty\}$ 
for each $i\le r$.}

\noindent We have\cmmnt{ moreover} the following formula for the
measures of such sets, writing $\mu$ for Lebesgue measure:

\Centerline{$\mu\ooint{a,b}=\mu[a,b]=\prod_{i=1}^r\beta_i-\alpha_i$}

\noindent whenever $a\le b$ in $\BbbR^r$.   Consequently every
countable subset of $\BbbR^r$ is measurable and of zero measure.

\proof{{\bf (a)} I show first that all open subsets of $\BbbR^r$
are measurable.   \Prf\ Let $G\subseteq\BbbR^r$ be open.   Let
$K\subseteq\BbbQ^r\times\BbbQ^r$ be the set of pairs $(c,d)$ of
$r$-tuples of rational
numbers such that $\coint{c,d}\subseteq G$.   Now by the remarks in
111E-111F
-- specifically, 111Eb, showing that $\Bbb Q$ is countable, 111F(b-iii),
showing that the product of two countable sets is countable, and
111F(b-i), showing that subsets of countable sets are countable -- we
see, inducing
on $r$, that $\BbbQ^r$ is countable, and that $K$
is countable.   Also, every $\coint{c,d}$ is measurable, being

\Centerline{$\bigcap_{i\le r}H_{i\delta_i}\setminus H_{i\gamma_i}$,}

\noindent in the language of 115F, if $c=(\gamma_1,\ldots,\gamma_r)$ and
$d=(\delta_1,\ldots,\delta_r)$.   So, by 111Fa,
$G'=\bigcup_{(r,s)\in K}\coint{r,s}$ is measurable.

By the definition of $K$, $G'\subseteq G$.   On the other hand, if
$x\in G$, there is an $\epsilon>0$ such that $y\in G$ whenever
$\|y-x\|<\epsilon$.
Now for each $i$ there are rational numbers $\gamma_i$, $\delta_i$ such
that $\gamma_i\le\xi_i<\delta_i$ and
$\delta_i-\gamma_i\le\Bover{\epsilon}{\sqrt r}$.   If $y\in\coint{c,d}$
then $|\eta_i-\xi_i|<\Bover{\epsilon}{\sqrt r}$ for every $i$ so
$\|y-x\|<\epsilon$ and $y\in G$.   Accordingly $(c,d)\in K$ and
$x\in\coint{c,d}\subseteq G'$.
As $x$ is arbitrary, $G=G'$ and $G$ is measurable.\ \Qed

\medskip

{\bf (b)} Now the family $\Sigma$ of Lebesgue measurable sets is a
$\sigma$-algebra of subsets of $\BbbR^r$ including the family of open
sets, so must contain every Borel set, by the definition of Borel set
(111G).

\medskip

{\bf (c)} Of the types of interval considered, all the open intervals
are actually open sets, so are surely Borel.   
A closed interval $[a,b]$ is expressible as the intersection
$\bigcap_{n\in\Bbb N}\ooint{a-2^{-n}\tbf{1},b+2^{-n}\tbf{1}}$ 
of a sequence of open intervals, so is Borel.

\medskip

{\bf (d)} To compute the measures, we already know from 115Db that
$\mu\coint{a,b}=\prod_{i=1}^r\beta_i-\alpha_i$ if $a\le b$ in $\BbbR^r$.   
For the other types of bounded
interval, it is enough to note that if $-\infty<\alpha_i<\beta_i<\infty$
for every $i$, then

\Centerline{$\coint{a+\epsilon\tbf{1},b}
\subseteq\ooint{a,b}\subseteq[a,b]\subseteq
\coint{a,b+\epsilon\tbf{1}}$}

\noindent whenever $\epsilon>0$ in $\Bbb R$.   So

\Centerline{$\mu\ooint{a,b}\le\mu[a,b]
\le\inf_{\epsilon>0}\mu\coint{a,b+\epsilon\tbf{1}}
=\inf_{\epsilon>0}\prod_{i=1}^r(\beta_i-\alpha_i+\epsilon)
=\prod_{i=1}^r\beta_i-\alpha_i$.}

\noindent If $\beta_i=\alpha_i$ for any $i$, then we must have

\Centerline{$\mu\ooint{a,b}=\mu[a,b]=
0=\prod_{i=1}^r\beta_i-\alpha_i$.}

\noindent If $\beta_i>\alpha_i$ for every $i$, then set
$\epsilon_0=\min_{i\le r}\beta_i-\alpha_i>0$;  then

$$\eqalign{\mu[a,b]\ge\mu\ooint{a,b}
&\ge\sup_{0<\epsilon\le\epsilon_0}\mu\coint{a+\epsilon\tbf{1},b}\cr
&=\sup_{0<\epsilon\le\epsilon_0}
  \prod_{i=1}^r(\beta_i-\alpha_i-\epsilon)
=\prod_{i=1}^r\beta_i-\alpha_i.\cr}$$

\noindent So in this case

\Centerline{$\prod_{i=1}^r\beta_i-\alpha_i\le\mu\ooint{a,b}
\le\mu[a,b]\le\prod_{i=1}^r\beta_i-\alpha_i$}

\noindent and

\Centerline{$\mu\ooint{a,b}
=\mu[a,b]=\prod_{i=1}^r\beta_i-\alpha_i$.}

\medskip

{\bf (e)} By (d), $\mu\{a\}=\mu[a,a]=0$ for every $a\in\BbbR^r$.   If
$A\subseteq\BbbR^r$ is countable, it is either empty or expressible as
$\{a_n:n\in\Bbb N\}$.   In the former case $\mu A=\mu\emptyset=0$;  in
the latter, $A=\bigcup_{n\in\Bbb N}\{a_n\}$ is Borel and $\mu
A\le\sum_{n=0}^{\infty}\mu\{a_n\}=0$.
}%end of proof of 115G

\exercises{
\leader{115X}{Basic exercises}  If you skipped \S114, you should now
return to 114X and assure yourself that you can do the exercises there as
well as those below.

\header{115Xa}{\bf (a)} Show that if $I$, $J$ are half-open intervals in
$\BbbR^r$, then $I\setminus J$ is expressible as the union of at most
$2r$
disjoint half-open intervals.   Hence show that (i) any finite union of
half-open intervals is expressible as a finite union of disjoint
half-open intervals (ii) any countable union of half-open intervals is
expressible as the union of a disjoint sequence of half-open intervals.

\sqheader 115Xb Write $\theta$ for Lebesgue outer measure, $\mu$
for Lebesgue
measure on $\BbbR^r$.   Show that $\theta A=\inf\{\mu E:E$ is Lebesgue
measurable, $A\subseteq E\}$ for every $A\subseteq\BbbR^r$.
\Hint{consider sets $E$ of the form $\bigcup_{j\in\Bbb N}I_j$, where
$\sequence{j}{I_j}$ is a sequence of half-open intervals.}

\header{115Xc}{\bf (c)} Let $E\subseteq\BbbR^r$ be a set of finite
measure for Lebesgue
measure $\mu$.   Show that for every $\epsilon>0$ there is a disjoint
family $I_0,\ldots,I_n$ of half-open intervals such that
$\mu(E\symmdiff\bigcup_{j\le n}I_j)\le\epsilon$.   \Hint{let
$\langle J_j\rangle_{j\in\Bbb N}$ be a sequence of half-open intervals
such that $E\subseteq\bigcup_{j\in\Bbb N}J_j$ and
$\sum_{j=0}^{\infty}\mu J_j\le\mu E+\bover{1}{2}\epsilon$.   Now take a
suitably large $m$ and express $\bigcup_{j\le m}J_j$ as a disjoint union
of half-open intervals.}

\sqheader 115Xd Suppose that $c\in\BbbR^r$.   (i) Show that
$\theta(A+c)=\theta A$ for every $A\subseteq\BbbR^r$, where
$A+c=\{x+c:x\in A\}$.   (ii)
Show that if $E\subseteq \BbbR^r$ is measurable
so is $E+c$, and that in this case $\mu(E+c)=\mu E$.

\header{115Xe}{\bf (e)} Suppose that $\gamma>0$.   (i) Show that
$\theta(\gamma A)=\gamma^r\theta A$ for every $A\subseteq\BbbR^r$, where
$\gamma A=\{\gamma x:x\in A\}$.   (ii) Show that if $E\subseteq \BbbR^r$ is
measurable so is $\gamma E$, and that in this case
$\mu(\gamma E)=\gamma^r\mu E$

\vleader{48pt}{115Y}{Further exercises (a)}
%\spheader 115Ya
(i) Suppose that $M$ is a strictly positive integer and $k_i$, $l_i$ are
integers for $1\le i\le r$.   Set $\alpha_i=k_i/M$ and
$\beta_i=l_i/M$ for each $i$, and $I=\coint{a,b}$.   Show that
$\lambda I=\#(J)/M^r$, where $J$ is $\{z:z\in\BbbZ^r$, $\bover1Mz\in I\}$.
(ii) Show that if a
half-open interval $I\subseteq\BbbR^r$ is covered by a {\it finite}
sequence $I_0,\ldots,I_m$ of half-open intervals, and all the coordinates
involved in specifying the intervals $I$, $I_0,\ldots,I_m$ are rational,
then $\lambda I\le\sum_{j=0}^m\lambda I_j$.   (iii) Assuming the
Heine-Borel theorem in the form

\inset{\noindent
whenever $[a,b]$ is a closed interval in $\BbbR^r$ which is
covered by a sequence $\sequence{j}{\ooint{a^{(j)},b^{(j)}}}$ of open
intervals, there is an $m\in\Bbb N$ such that
$[a,b]\subseteq\bigcup_{j\le m}\ooint{a^{(j)},b^{(j)}}$,}

\noindent prove 115B.   \Hint{if
$\coint{a,b}\subseteq\bigcup_{j\in\Bbb N}\coint{a^{(j)},b^{(j)}}$,
replace $\coint{a,b}$ by a smaller closed
interval and each $\coint{a^{(j)},b^{(j)}}$ by a larger open interval,
changing the volumes by adequately small amounts.}
%{\smc Facenda \& Freniche 02}

\spheader 115Yb(i)
Show that if $A\subseteq\BbbR^r$ and $\epsilon>0$, there is
an open set $G\supseteq A$ such that $\theta G\le\theta A+\epsilon$,
where $\theta$ is Lebesgue outer measure.   (ii) Show that if
$E\subseteq\BbbR^r$ is Lebesgue measurable and $\epsilon>0$, there is an
open set $G\supseteq E$ such that $\mu(G\setminus E)\le\epsilon$, where
$\mu$ is Lebesgue measure.   \Hint{consider first the case of
bounded $E$.}   (iii) Show that if $E\subseteq\BbbR^r$ is Lebesgue
measurable, there are Borel sets $H_1$, $H_2$ such that $H_1\subseteq
E\subseteq H_2$ and $\mu(H_2\setminus E)=\mu(E\setminus H_1)=0$.
\Hint{use (ii) to find $H_2$, and then consider the complement of
$E$.}

\spheader 115Yc Write $\theta$ for Lebesgue outer measure on
$\BbbR^r$.   Show
that a set $E\subseteq\BbbR^r$ is Lebesgue measurable iff
$\theta([-\tbf{n},\tbf{n}]\cap E)+\theta([-\tbf{n},\tbf{n}]\setminus
E)=(2n)^r$ for every $n\in\Bbb N$, writing $\tbf{n}=(n,\ldots,n)$.   \Hint{use 115Yb to show that
for each $n$ there are
measurable sets $F_n$, $H_n$ such that
$F_n\subseteq[-\tbf{n},\tbf{n}]\cap
E\subseteq H_n$ and $H_n\setminus F_n$ is negligible.}

\header{115Yd}{\bf (d)} Assuming that there is a set $A\subseteq\Bbb R$
which is not a Borel set, show that there is a family $\Cal E$ of
half-open intervals in $\BbbR^2$ such that $\bigcup\Cal E$ is not a
Borel set.   \Hint{consider $\Cal
E=\{\coint{\xi,1+\xi}\times\coint{-\xi,1-\xi}:\xi\in A\}$.}

\header{115Ye}{\bf (e)} Let $X$ be a set and $\Cal A$ a {\bf semiring} of
subsets of $X$, that is, a family of subsets of $X$ such that

\inset{$\emptyset\in\Cal A$,}

\inset{$E\cap F\in\Cal A$ for all $E$, $F\in\Cal A$,}

\inset{whenever $E$, $F\in\Cal A$ there are disjoint
$E_0,\ldots,E_n\in\Cal A$ such that $E\setminus F=E_0\cup\ldots\cup
E_n$.}

\noindent Let $\lambda:\Cal A\to[0,\infty]$ be a functional such that

\inset{$\lambda\emptyset=0$,}

\inset{$\lambda E=\sum_{i=0}^{\infty}\lambda E_i$ whenever $E\in\Cal A$
and $\sequence{i}{E_i}$ is a disjoint sequence in $\Cal A$ with union
$E$.}

\noindent Show that there is a measure $\mu$ on $X$ extending $\lambda$.
\Hint{use the method of 113Yi.}
}%end of exercises

\endnotes{
\Notesheader{115} In the notes to \S114 I ran over the methods so far
available to us for the construction of measure spaces.   To the list
there we can now add Lebesgue measure on $\BbbR^r$.

If you look back at \S114, you will see that I have deliberately copied
the exposition there.   I hope that this duplication will help you to see
the essential elements of the method, which are three:  a primitive
concept of volume (114A/115A);  countable subadditivity (114B/115B);  and
measurability of building blocks (114F/115F).

Concerning the `primitive concept of volume' there is not much to be
said.   The ideas of length of an interval, area of a rectangle and
volume of a cuboid go back to the beginning of mathematics.   I use
`half-open intervals', as defined in 114Aa/115Ab, for purely technical
reasons, because they fit together neatly (see 115Xa and 115Ye);
if we started with `open' or
`closed' intervals the method would still work.   One thing is perhaps
worth mentioning:  the blocks I use are all upright, with edges parallel
to the coordinate axes.   It is in fact a non-trivial exercise to prove
that a block in any other orientation has the right Lebesgue measure, and
I delay this until Chapter 26.   For the moment we are looking for the
shortest safe path to a precise definition, and the fact that rotating a
set doesn't change its Lebesgue measure will have to wait.

The big step is `countable subadditivity':  the fact that if one block is
covered by a sequence of other blocks, its volume is less than or equal
to the sum of theirs.   This is surely necessary if blocks are to be
measurable with the right measures, by 112Cd.   (What is remarkable is
that it is so nearly sufficient.)   Here we have some work to do, and in
the $r$-dimensional case there is a substantial hill to climb.   You can
do the climb in two stages if you look up the Heine-Borel theorem
(115Ya);  but as I try to explain in the remarks following 115B, I do not
think that this route avoids any of the real difficulties.

The third thing we must check is that blocks are measurable in the
technical sense described by \Caratheodory's theorem.   This is because
they are obtainable by the operations of intersection and union and
complementation from half-spaces, and half-spaces are measurable for very
straightforward reasons (114F/115F).   Now we are well away, and I do
very little more, only checking that open sets, and therefore Borel sets,
are measurable, and that closed and open intervals have the right
measures (114G/115G).   Some more properties of Lebesgue measure can be
found in \S134.   But every volume, if not quite every chapter, of this
treatise will introduce further features of this extraordinary
construction.
}%end of notes

\frnewpage


\frfilename{mt11conc.tex}
\versiondate{19.11.15}
\copyrightdate{2000}

\def\chaptername{Concordance}
\def\sectionname{Concordance}

\centerline{\bf Concordance for Chapter 11}

\bigskip

I list here the section and paragraph numbers which have 
(to my knowledge) appeared in print in references to this chapter,
and which have since been changed.

\medskip

\leaveitout{\leader{1{}12D}{Negligible sets} This has been referred to in
{\smc Okada \& Ricker 19}.

\leader{1{}12C, 1{}12E}{Measurable functions} These have been referred to in
{\smc Okada \& Ricker 19}.}
 
\leader{1{}12E-1{}12F}{Image measures} These paragraphs,
referred to in the 2001 and 2003 editions of Volume 2, and the 2003 and
2006 editions of Volume 4, have been moved to 234C-234D in Volume 2.

\leader{1{}12Ya}{Sums of measures} This material, referred to in the 2001
and 2003 editions of Volume 2, has been moved to 234G in Volume 2.


\discrpage


\references{
\frfilename{mt1r.tex}
\versiondate{31.5.03}

% except for references in ...X., ...Y., ...Z. give locations in form
%  342L  rather than  342Lb

\gdef\topparagraph{}
\gdef\bottomparagraph{}
\gdef\newparagraph{}
\gdef\chaptername{References}
\gdef\sectionname{References}
\gdef\headlinesectionname{References}

\centerline{\bf References for Volume 1}

\medskip

In addition to those (very few) works which I have mentioned in the
course of this volume, I list some of the books from which I myself
learnt measure theory, as a mark of grateful respect, and to
give you an opportunity to sample alternative approaches.

\medskip

Bartle R.G.\ [66] {\it The Elements of Integration.}  Wiley, 1966.

Cohn D.L.\ [80] {\it Measure Theory.}  Birkh\"auser, 1980.

Facenda J.A. \& Freniche F.J.\ [02]
{\it Integraci\'on de funciones de varias variables.}
Ediciones Pir\'amide, 2002.

Halmos P.R.\ [50] {\it Measure Theory.}   Van Nostrand, 1950.

Lebesgue H.\ [1901] `Sur une g\'en\'eralisation de l'int\'egrale
d\'efinie', C.R.Acad.Sci.\ (Paris) 132 (1901) 1025-1027.
Reprinted in {\smc Lebesgue 72}, vol.\ 1.

\leaveitout{Lebesgue H.\ [1902] {\it Int\'egrale, Longueur, Aire,}  th\`ese pr\'esent\'ee \`a la Facult\'e de Sciences de Paris,  Annali di Matematica Pura Appl.\ 7 (1902).   Reprinted in {\smc
Lebesgue 72}, vol.\ 1.}

Lebesgue H.\ [1904] {\it Le\c{c}ons sur l'int\'egration et la recherche
des fonctions primitives.} Gauthier-Villars, 1904.
Reprinted in {\smc Lebesgue 72}, vol.\ 2.
\cmmnt{[Vol.\ 1 {\it intro.}]}

Lebesgue H.\ [1918] `Remarques sur les th\'eories de la mesure et de l'int\'egration', Ann.\ Sci.\ Ecole Norm.\ Sup.\ 35 (1918) 191-250.
Reprinted in {\smc Lebesgue 72}, vol.\ 2.
\cmmnt{[Vol.\ 1 {\it intro.}]}

Lebesgue H.\ [72] {\it Oeuvres Scientifiques.}   L'Enseignement Math\'ematique, Institut de Math\'ematiques, Univ.\ de Gen\`eve, 1972.

Munroe M.E.\ [53] {\it Introduction to Measure and Integration.}
Addison-Wesley, 1953.

Royden H.L.\ [63] {\it Real Analysis.}  Macmillan, 1963.

Solovay R.M.\ [70] `A model of set theory in which every set of reals
is Lebesgue measurable', Annals of Math.\ 92 (1970) 1-56.   \cmmnt{[134C.]}

Tao T.\ [11] {\it An introduction to measure theory.}   Amer.\ Math.\ Soc.,
2011;   {\tt terrytao.files.wordpress.com/2011/01/measure-book1.pdf}.

Widom H.\ [69] {\it Lectures on Measure and Integration.}  Van Nostrand Reinhold, 1969.

Williamson J.H.\ [62] {\it Lebesgue Integration.}  Holt, Rinehart \& Winston, 1962.

\frnewpage

}%end of references

%---------------------------------------------------------------%

\end



